Did I just discover this integration formula?












39














One night, I discovered an integration relationship. That relationship allows to quickly integrate squares of functions (and even more, but I will talk about this at the end).



I was wondering if anyone has found a formula like this before. So I researched on the internet, but couldn't find anything like this.





The formula:



$$int f(x)^2,dx;=;xf(x)^2;-;2f(x)cdot F^{-1}_{(1)}(f(x));+;2,cdot F^{-1}_{(2)}(f(x));+;C$$



Where $F^{-1}_{(n)}$ denotes the $n$th anti-derivative of the inverse function of $f$.



The formula looks rather complicated, but it's really not, on further inspection. Note: the formula doesn't work on the function $f(x)=x$ for a reason that I wasn't able to determine yet. Edit: it actually works.





An example in action:



Let's compute $int ln^2x,dx$. We have then:



$f(x)=ln x$



$f^{-1}(x)=e^x$



$F^{-1}_{(1)}(x)=e^x$



$F^{-1}_{(2)}(x)=e^x$



Applying the formula, the integral becomes:



$$intln^2x,dx=xln^2x-2ln xcdot e^{ln x}+2cdot e^{ln x}+C$$
$$=xln^2x-2xln x+2x+C$$





Derivation (for the curious):



I derived this formula by substituting for inverse functions and doing repeated integration by parts.



First, substitute $x=f^{-1}(u)$. Then, we have $dx=df^{-1}(u)$. This changes the original integral to:



$$int f(x)^2,dx=int f(f^{-1}(u))^2,df^{-1}(u)=int u^2,df^{-1}(u)$$



At this point, I did integration by parts (probably the funkiest integration by parts you have ever seen). Note, that I am using the "DI table" trick for integration by parts, where in one column, derivatives of one function are specified, and integrals of another are put into the other column. Terms are multiplied diagonally left-down.



$$begin{array}{ l | c | r }
pm & D & I \
hline
+ & u^2 & df^{-1}(u) \
- & 2u & f^{-1}(u) \
+ & 2 & F^{-1}_{(1)}(u) \
- & 0 & F^{-1}_{(2)}(u) \
end{array}$$



$$int u^2,df^{-1}(u);=;u^2f^{-1}(u)-2uF^{-1}_{(1)}(u)+2F^{-1}_{(2)}(u)+C$$



Now, simply subsitute back $u=f(x)$ and we arrive at the formula.





Generalization to composition of functions:



It didn't take long for me to realize that this method can be extended to compositions of functions. Just for the curious folks, this is my integral of composition formula:



$$int f(g(x)),dx;=;sum^{infty}_{k=0}(-1)^kcdot D_k(g(x))cdot A_k(g(x)) + C$$



Where $D_k={d^k fover dx^k}$ and $A_k={d^{-k}g^{-1}over dx^{-k}}$.



Interesting, isn't it?





Back to the question:



Have I discovered this formula? If I didn't can someone point me to a further reading on this topic? I really can't find anything myself, probably because I don't know how to search.










share|cite|improve this question




















  • 3




    at a first glance it seems correct. However I doubt that nobody had written a similar formula in the previous centuries :)
    – Masacroso
    Dec 9 '18 at 4:24






  • 3




    I'm pretty sure this does work for $f(x)=x$
    – WW1
    Dec 9 '18 at 5:20










  • @WW1 oh yeah, it does :) I don't know how it didn't work for me before
    – KKZiomek
    Dec 9 '18 at 7:14
















39














One night, I discovered an integration relationship. That relationship allows to quickly integrate squares of functions (and even more, but I will talk about this at the end).



I was wondering if anyone has found a formula like this before. So I researched on the internet, but couldn't find anything like this.





The formula:



$$int f(x)^2,dx;=;xf(x)^2;-;2f(x)cdot F^{-1}_{(1)}(f(x));+;2,cdot F^{-1}_{(2)}(f(x));+;C$$



Where $F^{-1}_{(n)}$ denotes the $n$th anti-derivative of the inverse function of $f$.



The formula looks rather complicated, but it's really not, on further inspection. Note: the formula doesn't work on the function $f(x)=x$ for a reason that I wasn't able to determine yet. Edit: it actually works.





An example in action:



Let's compute $int ln^2x,dx$. We have then:



$f(x)=ln x$



$f^{-1}(x)=e^x$



$F^{-1}_{(1)}(x)=e^x$



$F^{-1}_{(2)}(x)=e^x$



Applying the formula, the integral becomes:



$$intln^2x,dx=xln^2x-2ln xcdot e^{ln x}+2cdot e^{ln x}+C$$
$$=xln^2x-2xln x+2x+C$$





Derivation (for the curious):



I derived this formula by substituting for inverse functions and doing repeated integration by parts.



First, substitute $x=f^{-1}(u)$. Then, we have $dx=df^{-1}(u)$. This changes the original integral to:



$$int f(x)^2,dx=int f(f^{-1}(u))^2,df^{-1}(u)=int u^2,df^{-1}(u)$$



At this point, I did integration by parts (probably the funkiest integration by parts you have ever seen). Note, that I am using the "DI table" trick for integration by parts, where in one column, derivatives of one function are specified, and integrals of another are put into the other column. Terms are multiplied diagonally left-down.



$$begin{array}{ l | c | r }
pm & D & I \
hline
+ & u^2 & df^{-1}(u) \
- & 2u & f^{-1}(u) \
+ & 2 & F^{-1}_{(1)}(u) \
- & 0 & F^{-1}_{(2)}(u) \
end{array}$$



$$int u^2,df^{-1}(u);=;u^2f^{-1}(u)-2uF^{-1}_{(1)}(u)+2F^{-1}_{(2)}(u)+C$$



Now, simply subsitute back $u=f(x)$ and we arrive at the formula.





Generalization to composition of functions:



It didn't take long for me to realize that this method can be extended to compositions of functions. Just for the curious folks, this is my integral of composition formula:



$$int f(g(x)),dx;=;sum^{infty}_{k=0}(-1)^kcdot D_k(g(x))cdot A_k(g(x)) + C$$



Where $D_k={d^k fover dx^k}$ and $A_k={d^{-k}g^{-1}over dx^{-k}}$.



Interesting, isn't it?





Back to the question:



Have I discovered this formula? If I didn't can someone point me to a further reading on this topic? I really can't find anything myself, probably because I don't know how to search.










share|cite|improve this question




















  • 3




    at a first glance it seems correct. However I doubt that nobody had written a similar formula in the previous centuries :)
    – Masacroso
    Dec 9 '18 at 4:24






  • 3




    I'm pretty sure this does work for $f(x)=x$
    – WW1
    Dec 9 '18 at 5:20










  • @WW1 oh yeah, it does :) I don't know how it didn't work for me before
    – KKZiomek
    Dec 9 '18 at 7:14














39












39








39


18





One night, I discovered an integration relationship. That relationship allows to quickly integrate squares of functions (and even more, but I will talk about this at the end).



I was wondering if anyone has found a formula like this before. So I researched on the internet, but couldn't find anything like this.





The formula:



$$int f(x)^2,dx;=;xf(x)^2;-;2f(x)cdot F^{-1}_{(1)}(f(x));+;2,cdot F^{-1}_{(2)}(f(x));+;C$$



Where $F^{-1}_{(n)}$ denotes the $n$th anti-derivative of the inverse function of $f$.



The formula looks rather complicated, but it's really not, on further inspection. Note: the formula doesn't work on the function $f(x)=x$ for a reason that I wasn't able to determine yet. Edit: it actually works.





An example in action:



Let's compute $int ln^2x,dx$. We have then:



$f(x)=ln x$



$f^{-1}(x)=e^x$



$F^{-1}_{(1)}(x)=e^x$



$F^{-1}_{(2)}(x)=e^x$



Applying the formula, the integral becomes:



$$intln^2x,dx=xln^2x-2ln xcdot e^{ln x}+2cdot e^{ln x}+C$$
$$=xln^2x-2xln x+2x+C$$





Derivation (for the curious):



I derived this formula by substituting for inverse functions and doing repeated integration by parts.



First, substitute $x=f^{-1}(u)$. Then, we have $dx=df^{-1}(u)$. This changes the original integral to:



$$int f(x)^2,dx=int f(f^{-1}(u))^2,df^{-1}(u)=int u^2,df^{-1}(u)$$



At this point, I did integration by parts (probably the funkiest integration by parts you have ever seen). Note, that I am using the "DI table" trick for integration by parts, where in one column, derivatives of one function are specified, and integrals of another are put into the other column. Terms are multiplied diagonally left-down.



$$begin{array}{ l | c | r }
pm & D & I \
hline
+ & u^2 & df^{-1}(u) \
- & 2u & f^{-1}(u) \
+ & 2 & F^{-1}_{(1)}(u) \
- & 0 & F^{-1}_{(2)}(u) \
end{array}$$



$$int u^2,df^{-1}(u);=;u^2f^{-1}(u)-2uF^{-1}_{(1)}(u)+2F^{-1}_{(2)}(u)+C$$



Now, simply subsitute back $u=f(x)$ and we arrive at the formula.





Generalization to composition of functions:



It didn't take long for me to realize that this method can be extended to compositions of functions. Just for the curious folks, this is my integral of composition formula:



$$int f(g(x)),dx;=;sum^{infty}_{k=0}(-1)^kcdot D_k(g(x))cdot A_k(g(x)) + C$$



Where $D_k={d^k fover dx^k}$ and $A_k={d^{-k}g^{-1}over dx^{-k}}$.



Interesting, isn't it?





Back to the question:



Have I discovered this formula? If I didn't can someone point me to a further reading on this topic? I really can't find anything myself, probably because I don't know how to search.










share|cite|improve this question















One night, I discovered an integration relationship. That relationship allows to quickly integrate squares of functions (and even more, but I will talk about this at the end).



I was wondering if anyone has found a formula like this before. So I researched on the internet, but couldn't find anything like this.





The formula:



$$int f(x)^2,dx;=;xf(x)^2;-;2f(x)cdot F^{-1}_{(1)}(f(x));+;2,cdot F^{-1}_{(2)}(f(x));+;C$$



Where $F^{-1}_{(n)}$ denotes the $n$th anti-derivative of the inverse function of $f$.



The formula looks rather complicated, but it's really not, on further inspection. Note: the formula doesn't work on the function $f(x)=x$ for a reason that I wasn't able to determine yet. Edit: it actually works.





An example in action:



Let's compute $int ln^2x,dx$. We have then:



$f(x)=ln x$



$f^{-1}(x)=e^x$



$F^{-1}_{(1)}(x)=e^x$



$F^{-1}_{(2)}(x)=e^x$



Applying the formula, the integral becomes:



$$intln^2x,dx=xln^2x-2ln xcdot e^{ln x}+2cdot e^{ln x}+C$$
$$=xln^2x-2xln x+2x+C$$





Derivation (for the curious):



I derived this formula by substituting for inverse functions and doing repeated integration by parts.



First, substitute $x=f^{-1}(u)$. Then, we have $dx=df^{-1}(u)$. This changes the original integral to:



$$int f(x)^2,dx=int f(f^{-1}(u))^2,df^{-1}(u)=int u^2,df^{-1}(u)$$



At this point, I did integration by parts (probably the funkiest integration by parts you have ever seen). Note, that I am using the "DI table" trick for integration by parts, where in one column, derivatives of one function are specified, and integrals of another are put into the other column. Terms are multiplied diagonally left-down.



$$begin{array}{ l | c | r }
pm & D & I \
hline
+ & u^2 & df^{-1}(u) \
- & 2u & f^{-1}(u) \
+ & 2 & F^{-1}_{(1)}(u) \
- & 0 & F^{-1}_{(2)}(u) \
end{array}$$



$$int u^2,df^{-1}(u);=;u^2f^{-1}(u)-2uF^{-1}_{(1)}(u)+2F^{-1}_{(2)}(u)+C$$



Now, simply subsitute back $u=f(x)$ and we arrive at the formula.





Generalization to composition of functions:



It didn't take long for me to realize that this method can be extended to compositions of functions. Just for the curious folks, this is my integral of composition formula:



$$int f(g(x)),dx;=;sum^{infty}_{k=0}(-1)^kcdot D_k(g(x))cdot A_k(g(x)) + C$$



Where $D_k={d^k fover dx^k}$ and $A_k={d^{-k}g^{-1}over dx^{-k}}$.



Interesting, isn't it?





Back to the question:



Have I discovered this formula? If I didn't can someone point me to a further reading on this topic? I really can't find anything myself, probably because I don't know how to search.







calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 7:15

























asked Dec 9 '18 at 4:20









KKZiomek

2,0611339




2,0611339








  • 3




    at a first glance it seems correct. However I doubt that nobody had written a similar formula in the previous centuries :)
    – Masacroso
    Dec 9 '18 at 4:24






  • 3




    I'm pretty sure this does work for $f(x)=x$
    – WW1
    Dec 9 '18 at 5:20










  • @WW1 oh yeah, it does :) I don't know how it didn't work for me before
    – KKZiomek
    Dec 9 '18 at 7:14














  • 3




    at a first glance it seems correct. However I doubt that nobody had written a similar formula in the previous centuries :)
    – Masacroso
    Dec 9 '18 at 4:24






  • 3




    I'm pretty sure this does work for $f(x)=x$
    – WW1
    Dec 9 '18 at 5:20










  • @WW1 oh yeah, it does :) I don't know how it didn't work for me before
    – KKZiomek
    Dec 9 '18 at 7:14








3




3




at a first glance it seems correct. However I doubt that nobody had written a similar formula in the previous centuries :)
– Masacroso
Dec 9 '18 at 4:24




at a first glance it seems correct. However I doubt that nobody had written a similar formula in the previous centuries :)
– Masacroso
Dec 9 '18 at 4:24




3




3




I'm pretty sure this does work for $f(x)=x$
– WW1
Dec 9 '18 at 5:20




I'm pretty sure this does work for $f(x)=x$
– WW1
Dec 9 '18 at 5:20












@WW1 oh yeah, it does :) I don't know how it didn't work for me before
– KKZiomek
Dec 9 '18 at 7:14




@WW1 oh yeah, it does :) I don't know how it didn't work for me before
– KKZiomek
Dec 9 '18 at 7:14










2 Answers
2






active

oldest

votes


















12














Maybe I'm missing something, but it seems you're just making the substitution $u = f(x)$ in the $f(x)^2$ case and $u = g(x)$ in the $f(g(x))$ case and then integrating by parts (multiple times). This can certainly be useful depending on the functional forms of $f$ and $g$.






share|cite|improve this answer

















  • 1




    Yes, it's just a u=f(x) substitution. Pretty simple, but allows for turning a composition of general functions into a product of a specific function and a differential. This simplicty raises more probability that someone discovered it before just in some obscure paper or in a way that is hard to search online. So I'll just assume I rediscovered it.
    – KKZiomek
    Dec 9 '18 at 15:32





















10














My first thought was that if this were true, I'd have seen it already in conjunction with the tables of integrals in one of the standard reference books.
But it seems it is true.



The reason the formula isn't more prominent may be because it requires $f(x)$ to have an inverse, and then it requires that inverse to be integrated twice.
Those requirements put a lot of functions out of reach of this formula.
For example, consider $f(x) = xe^{x^2/2},$ which is sometimes given as an example of why there (supposedly) is no general formula to integrate $f(x).$
(For example, see
how to calculate integral of square of a function
and Integral of Derivative squared).
Then $f^{-1}(x) = sqrt{mathrm W(x^2)},$ where W is the Lambert W function, but
(according to Wolfram Alpha) $sqrt{mathrm W(x^2)}$ has no integral expressible in elementary functions.



Out of curiosity, I worked this out for a few examples.
All of these are confirmed by other methods (which in these examples are
easier than the formula).





begin{align}
f(x) &= x \
f^{-1}(x) &= x \
F^{-1}_{(1)}(x) &= tfrac12 x^2 \
F^{-1}_{(2)}(x) &= tfrac16 x^3 \
F^{-1}_{(1)}(f(x)) &= tfrac12 x^2 \
F^{-1}_{(2)}(f(x)) &= tfrac16 x^3
end{align}



begin{align}
int x^2,dx
&= x cdot x^2 - 2x cdot tfrac12 x^2 + 2 cdot tfrac16 x^3 + C \
&= x^3 - x^3 + tfrac13 x^3 + C \
&= tfrac13 x^3 + C.
end{align}





begin{align}
f(x) &= sin x \
f^{-1}(x) &= arcsin x \
F^{-1}_{(1)}(x) &= sqrt{1 - x^2} + x arcsin x \
F^{-1}_{(2)}(x) &= tfrac14 (3x sqrt{1 - x^2} + 2 x^2 arcsin x + arcsin x) \
F^{-1}_{(1)}(f(x)) &= cos x + xsin x \
F^{-1}_{(2)}(f(x)) &= tfrac14 (3sin x cos x + 2 xsin^2 x + x)
end{align}



begin{align}
int sin^2 x,dx
&= x sin^2 x - 2sin x (cos x + xsin x)
+ 2 cdot tfrac14 (3sin x cos x + 2 xsin^2 x + x) + C \
&= x sin^2 x - 2sin x cos x - 2xsin^2 x
+ tfrac32sin x cos x + xsin^2 x + tfrac12x + C \
&= (x - 2x +x)sin^2 x + left(- 2 + tfrac32right)sin x cos x + tfrac12x + C \
&= -tfrac12 sin x cos x + tfrac12x + C. \
end{align}





begin{align}
f(x) &= e^x \
f^{-1}(x) &= ln x \
F^{-1}_{(1)}(x) &= x (ln x - 1) \
F^{-1}_{(2)}(x) &= tfrac14 x^2 (2 ln x - 3) \
F^{-1}_{(1)}(f(x)) &= e^x (x - 1) \
F^{-1}_{(2)}(f(x)) &= tfrac14 e^{2x} (2 x - 3)
end{align}



begin{align}
int (e^x)^2,dx
&= x e^{2x} - 2e^x cdot e^x (x - 1) + 2 cdot tfrac14 e^{2x} (2 x - 3) + C \
&= x e^{2x} - 2(x - 1)e^{2x} + left(x - tfrac32right) e^{2x} + C \
&= left(x - 2x + 2 + x - tfrac32right)e^{2x} + C \
&= tfrac12 e^{2x} + C. \
end{align}





begin{align}
f(x) &= e^{x^2} \
f^{-1}(x) &= sqrt{ln x} \
F^{-1}_{(1)}(x) &= xsqrt{ln x} - tfrac12sqrtpi,mathrm{erfi}(sqrt{ln x}) \
F^{-1}_{(2)}(x)
&= -tfrac12 sqrtpi x , mathrm{erfi}(sqrt{ln x})
+ tfrac18sqrt{2pi},mathrm{erfi}(sqrt{2ln x}) + tfrac12 x^2 sqrt{ln x} \
x f(x)^2 &= x e^{2x^2}, \
F^{-1}_{(1)}(f(x)) &= x e^{x^2} - tfrac12sqrtpi,mathrm{erfi}(x) \
2f(x)F^{-1}_{(1)}(f(x)) &= 2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x),\
2F^{-1}_{(2)}(f(x)) &= -sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} \
end{align}



begin{align}
int left(e^{x^2}right)^2,dx
&= x e^{2x^2} - left(2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x)right)
- sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} + C\
&= tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + C.
end{align}



This agrees with $int e^{u^2},du = tfrac12sqrtpi,mathrm{erfi}(u) + C$
with the substitution $u = sqrt2 x.$






share|cite|improve this answer

















  • 1




    Thank you for that analysis. So it seems that this formula works nicely only for certain functions, for others it is unnecessarily complicating. It seems that the best functions to integrate this method are powers of logarithms
    – KKZiomek
    Dec 9 '18 at 18:04






  • 1




    It did work out very nicely for $(ln x)^2$; no cancellation of unnecessary terms. I would not be surprised if there were some other functions it worked nicely for.
    – David K
    Dec 10 '18 at 0:35






  • 1




    True, but arguably $ln x $ is the *nicest* of all, as its inverse is $e^x$ whose antiderivtives are all $e^x$. Though, like you said, there may be functions that are nice enough for this method besides logarithms
    – KKZiomek
    Dec 10 '18 at 5:08











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032003%2fdid-i-just-discover-this-integration-formula%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









12














Maybe I'm missing something, but it seems you're just making the substitution $u = f(x)$ in the $f(x)^2$ case and $u = g(x)$ in the $f(g(x))$ case and then integrating by parts (multiple times). This can certainly be useful depending on the functional forms of $f$ and $g$.






share|cite|improve this answer

















  • 1




    Yes, it's just a u=f(x) substitution. Pretty simple, but allows for turning a composition of general functions into a product of a specific function and a differential. This simplicty raises more probability that someone discovered it before just in some obscure paper or in a way that is hard to search online. So I'll just assume I rediscovered it.
    – KKZiomek
    Dec 9 '18 at 15:32


















12














Maybe I'm missing something, but it seems you're just making the substitution $u = f(x)$ in the $f(x)^2$ case and $u = g(x)$ in the $f(g(x))$ case and then integrating by parts (multiple times). This can certainly be useful depending on the functional forms of $f$ and $g$.






share|cite|improve this answer

















  • 1




    Yes, it's just a u=f(x) substitution. Pretty simple, but allows for turning a composition of general functions into a product of a specific function and a differential. This simplicty raises more probability that someone discovered it before just in some obscure paper or in a way that is hard to search online. So I'll just assume I rediscovered it.
    – KKZiomek
    Dec 9 '18 at 15:32
















12












12








12






Maybe I'm missing something, but it seems you're just making the substitution $u = f(x)$ in the $f(x)^2$ case and $u = g(x)$ in the $f(g(x))$ case and then integrating by parts (multiple times). This can certainly be useful depending on the functional forms of $f$ and $g$.






share|cite|improve this answer












Maybe I'm missing something, but it seems you're just making the substitution $u = f(x)$ in the $f(x)^2$ case and $u = g(x)$ in the $f(g(x))$ case and then integrating by parts (multiple times). This can certainly be useful depending on the functional forms of $f$ and $g$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 8:43









maridia

1,06912




1,06912








  • 1




    Yes, it's just a u=f(x) substitution. Pretty simple, but allows for turning a composition of general functions into a product of a specific function and a differential. This simplicty raises more probability that someone discovered it before just in some obscure paper or in a way that is hard to search online. So I'll just assume I rediscovered it.
    – KKZiomek
    Dec 9 '18 at 15:32
















  • 1




    Yes, it's just a u=f(x) substitution. Pretty simple, but allows for turning a composition of general functions into a product of a specific function and a differential. This simplicty raises more probability that someone discovered it before just in some obscure paper or in a way that is hard to search online. So I'll just assume I rediscovered it.
    – KKZiomek
    Dec 9 '18 at 15:32










1




1




Yes, it's just a u=f(x) substitution. Pretty simple, but allows for turning a composition of general functions into a product of a specific function and a differential. This simplicty raises more probability that someone discovered it before just in some obscure paper or in a way that is hard to search online. So I'll just assume I rediscovered it.
– KKZiomek
Dec 9 '18 at 15:32






Yes, it's just a u=f(x) substitution. Pretty simple, but allows for turning a composition of general functions into a product of a specific function and a differential. This simplicty raises more probability that someone discovered it before just in some obscure paper or in a way that is hard to search online. So I'll just assume I rediscovered it.
– KKZiomek
Dec 9 '18 at 15:32













10














My first thought was that if this were true, I'd have seen it already in conjunction with the tables of integrals in one of the standard reference books.
But it seems it is true.



The reason the formula isn't more prominent may be because it requires $f(x)$ to have an inverse, and then it requires that inverse to be integrated twice.
Those requirements put a lot of functions out of reach of this formula.
For example, consider $f(x) = xe^{x^2/2},$ which is sometimes given as an example of why there (supposedly) is no general formula to integrate $f(x).$
(For example, see
how to calculate integral of square of a function
and Integral of Derivative squared).
Then $f^{-1}(x) = sqrt{mathrm W(x^2)},$ where W is the Lambert W function, but
(according to Wolfram Alpha) $sqrt{mathrm W(x^2)}$ has no integral expressible in elementary functions.



Out of curiosity, I worked this out for a few examples.
All of these are confirmed by other methods (which in these examples are
easier than the formula).





begin{align}
f(x) &= x \
f^{-1}(x) &= x \
F^{-1}_{(1)}(x) &= tfrac12 x^2 \
F^{-1}_{(2)}(x) &= tfrac16 x^3 \
F^{-1}_{(1)}(f(x)) &= tfrac12 x^2 \
F^{-1}_{(2)}(f(x)) &= tfrac16 x^3
end{align}



begin{align}
int x^2,dx
&= x cdot x^2 - 2x cdot tfrac12 x^2 + 2 cdot tfrac16 x^3 + C \
&= x^3 - x^3 + tfrac13 x^3 + C \
&= tfrac13 x^3 + C.
end{align}





begin{align}
f(x) &= sin x \
f^{-1}(x) &= arcsin x \
F^{-1}_{(1)}(x) &= sqrt{1 - x^2} + x arcsin x \
F^{-1}_{(2)}(x) &= tfrac14 (3x sqrt{1 - x^2} + 2 x^2 arcsin x + arcsin x) \
F^{-1}_{(1)}(f(x)) &= cos x + xsin x \
F^{-1}_{(2)}(f(x)) &= tfrac14 (3sin x cos x + 2 xsin^2 x + x)
end{align}



begin{align}
int sin^2 x,dx
&= x sin^2 x - 2sin x (cos x + xsin x)
+ 2 cdot tfrac14 (3sin x cos x + 2 xsin^2 x + x) + C \
&= x sin^2 x - 2sin x cos x - 2xsin^2 x
+ tfrac32sin x cos x + xsin^2 x + tfrac12x + C \
&= (x - 2x +x)sin^2 x + left(- 2 + tfrac32right)sin x cos x + tfrac12x + C \
&= -tfrac12 sin x cos x + tfrac12x + C. \
end{align}





begin{align}
f(x) &= e^x \
f^{-1}(x) &= ln x \
F^{-1}_{(1)}(x) &= x (ln x - 1) \
F^{-1}_{(2)}(x) &= tfrac14 x^2 (2 ln x - 3) \
F^{-1}_{(1)}(f(x)) &= e^x (x - 1) \
F^{-1}_{(2)}(f(x)) &= tfrac14 e^{2x} (2 x - 3)
end{align}



begin{align}
int (e^x)^2,dx
&= x e^{2x} - 2e^x cdot e^x (x - 1) + 2 cdot tfrac14 e^{2x} (2 x - 3) + C \
&= x e^{2x} - 2(x - 1)e^{2x} + left(x - tfrac32right) e^{2x} + C \
&= left(x - 2x + 2 + x - tfrac32right)e^{2x} + C \
&= tfrac12 e^{2x} + C. \
end{align}





begin{align}
f(x) &= e^{x^2} \
f^{-1}(x) &= sqrt{ln x} \
F^{-1}_{(1)}(x) &= xsqrt{ln x} - tfrac12sqrtpi,mathrm{erfi}(sqrt{ln x}) \
F^{-1}_{(2)}(x)
&= -tfrac12 sqrtpi x , mathrm{erfi}(sqrt{ln x})
+ tfrac18sqrt{2pi},mathrm{erfi}(sqrt{2ln x}) + tfrac12 x^2 sqrt{ln x} \
x f(x)^2 &= x e^{2x^2}, \
F^{-1}_{(1)}(f(x)) &= x e^{x^2} - tfrac12sqrtpi,mathrm{erfi}(x) \
2f(x)F^{-1}_{(1)}(f(x)) &= 2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x),\
2F^{-1}_{(2)}(f(x)) &= -sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} \
end{align}



begin{align}
int left(e^{x^2}right)^2,dx
&= x e^{2x^2} - left(2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x)right)
- sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} + C\
&= tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + C.
end{align}



This agrees with $int e^{u^2},du = tfrac12sqrtpi,mathrm{erfi}(u) + C$
with the substitution $u = sqrt2 x.$






share|cite|improve this answer

















  • 1




    Thank you for that analysis. So it seems that this formula works nicely only for certain functions, for others it is unnecessarily complicating. It seems that the best functions to integrate this method are powers of logarithms
    – KKZiomek
    Dec 9 '18 at 18:04






  • 1




    It did work out very nicely for $(ln x)^2$; no cancellation of unnecessary terms. I would not be surprised if there were some other functions it worked nicely for.
    – David K
    Dec 10 '18 at 0:35






  • 1




    True, but arguably $ln x $ is the *nicest* of all, as its inverse is $e^x$ whose antiderivtives are all $e^x$. Though, like you said, there may be functions that are nice enough for this method besides logarithms
    – KKZiomek
    Dec 10 '18 at 5:08
















10














My first thought was that if this were true, I'd have seen it already in conjunction with the tables of integrals in one of the standard reference books.
But it seems it is true.



The reason the formula isn't more prominent may be because it requires $f(x)$ to have an inverse, and then it requires that inverse to be integrated twice.
Those requirements put a lot of functions out of reach of this formula.
For example, consider $f(x) = xe^{x^2/2},$ which is sometimes given as an example of why there (supposedly) is no general formula to integrate $f(x).$
(For example, see
how to calculate integral of square of a function
and Integral of Derivative squared).
Then $f^{-1}(x) = sqrt{mathrm W(x^2)},$ where W is the Lambert W function, but
(according to Wolfram Alpha) $sqrt{mathrm W(x^2)}$ has no integral expressible in elementary functions.



Out of curiosity, I worked this out for a few examples.
All of these are confirmed by other methods (which in these examples are
easier than the formula).





begin{align}
f(x) &= x \
f^{-1}(x) &= x \
F^{-1}_{(1)}(x) &= tfrac12 x^2 \
F^{-1}_{(2)}(x) &= tfrac16 x^3 \
F^{-1}_{(1)}(f(x)) &= tfrac12 x^2 \
F^{-1}_{(2)}(f(x)) &= tfrac16 x^3
end{align}



begin{align}
int x^2,dx
&= x cdot x^2 - 2x cdot tfrac12 x^2 + 2 cdot tfrac16 x^3 + C \
&= x^3 - x^3 + tfrac13 x^3 + C \
&= tfrac13 x^3 + C.
end{align}





begin{align}
f(x) &= sin x \
f^{-1}(x) &= arcsin x \
F^{-1}_{(1)}(x) &= sqrt{1 - x^2} + x arcsin x \
F^{-1}_{(2)}(x) &= tfrac14 (3x sqrt{1 - x^2} + 2 x^2 arcsin x + arcsin x) \
F^{-1}_{(1)}(f(x)) &= cos x + xsin x \
F^{-1}_{(2)}(f(x)) &= tfrac14 (3sin x cos x + 2 xsin^2 x + x)
end{align}



begin{align}
int sin^2 x,dx
&= x sin^2 x - 2sin x (cos x + xsin x)
+ 2 cdot tfrac14 (3sin x cos x + 2 xsin^2 x + x) + C \
&= x sin^2 x - 2sin x cos x - 2xsin^2 x
+ tfrac32sin x cos x + xsin^2 x + tfrac12x + C \
&= (x - 2x +x)sin^2 x + left(- 2 + tfrac32right)sin x cos x + tfrac12x + C \
&= -tfrac12 sin x cos x + tfrac12x + C. \
end{align}





begin{align}
f(x) &= e^x \
f^{-1}(x) &= ln x \
F^{-1}_{(1)}(x) &= x (ln x - 1) \
F^{-1}_{(2)}(x) &= tfrac14 x^2 (2 ln x - 3) \
F^{-1}_{(1)}(f(x)) &= e^x (x - 1) \
F^{-1}_{(2)}(f(x)) &= tfrac14 e^{2x} (2 x - 3)
end{align}



begin{align}
int (e^x)^2,dx
&= x e^{2x} - 2e^x cdot e^x (x - 1) + 2 cdot tfrac14 e^{2x} (2 x - 3) + C \
&= x e^{2x} - 2(x - 1)e^{2x} + left(x - tfrac32right) e^{2x} + C \
&= left(x - 2x + 2 + x - tfrac32right)e^{2x} + C \
&= tfrac12 e^{2x} + C. \
end{align}





begin{align}
f(x) &= e^{x^2} \
f^{-1}(x) &= sqrt{ln x} \
F^{-1}_{(1)}(x) &= xsqrt{ln x} - tfrac12sqrtpi,mathrm{erfi}(sqrt{ln x}) \
F^{-1}_{(2)}(x)
&= -tfrac12 sqrtpi x , mathrm{erfi}(sqrt{ln x})
+ tfrac18sqrt{2pi},mathrm{erfi}(sqrt{2ln x}) + tfrac12 x^2 sqrt{ln x} \
x f(x)^2 &= x e^{2x^2}, \
F^{-1}_{(1)}(f(x)) &= x e^{x^2} - tfrac12sqrtpi,mathrm{erfi}(x) \
2f(x)F^{-1}_{(1)}(f(x)) &= 2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x),\
2F^{-1}_{(2)}(f(x)) &= -sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} \
end{align}



begin{align}
int left(e^{x^2}right)^2,dx
&= x e^{2x^2} - left(2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x)right)
- sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} + C\
&= tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + C.
end{align}



This agrees with $int e^{u^2},du = tfrac12sqrtpi,mathrm{erfi}(u) + C$
with the substitution $u = sqrt2 x.$






share|cite|improve this answer

















  • 1




    Thank you for that analysis. So it seems that this formula works nicely only for certain functions, for others it is unnecessarily complicating. It seems that the best functions to integrate this method are powers of logarithms
    – KKZiomek
    Dec 9 '18 at 18:04






  • 1




    It did work out very nicely for $(ln x)^2$; no cancellation of unnecessary terms. I would not be surprised if there were some other functions it worked nicely for.
    – David K
    Dec 10 '18 at 0:35






  • 1




    True, but arguably $ln x $ is the *nicest* of all, as its inverse is $e^x$ whose antiderivtives are all $e^x$. Though, like you said, there may be functions that are nice enough for this method besides logarithms
    – KKZiomek
    Dec 10 '18 at 5:08














10












10








10






My first thought was that if this were true, I'd have seen it already in conjunction with the tables of integrals in one of the standard reference books.
But it seems it is true.



The reason the formula isn't more prominent may be because it requires $f(x)$ to have an inverse, and then it requires that inverse to be integrated twice.
Those requirements put a lot of functions out of reach of this formula.
For example, consider $f(x) = xe^{x^2/2},$ which is sometimes given as an example of why there (supposedly) is no general formula to integrate $f(x).$
(For example, see
how to calculate integral of square of a function
and Integral of Derivative squared).
Then $f^{-1}(x) = sqrt{mathrm W(x^2)},$ where W is the Lambert W function, but
(according to Wolfram Alpha) $sqrt{mathrm W(x^2)}$ has no integral expressible in elementary functions.



Out of curiosity, I worked this out for a few examples.
All of these are confirmed by other methods (which in these examples are
easier than the formula).





begin{align}
f(x) &= x \
f^{-1}(x) &= x \
F^{-1}_{(1)}(x) &= tfrac12 x^2 \
F^{-1}_{(2)}(x) &= tfrac16 x^3 \
F^{-1}_{(1)}(f(x)) &= tfrac12 x^2 \
F^{-1}_{(2)}(f(x)) &= tfrac16 x^3
end{align}



begin{align}
int x^2,dx
&= x cdot x^2 - 2x cdot tfrac12 x^2 + 2 cdot tfrac16 x^3 + C \
&= x^3 - x^3 + tfrac13 x^3 + C \
&= tfrac13 x^3 + C.
end{align}





begin{align}
f(x) &= sin x \
f^{-1}(x) &= arcsin x \
F^{-1}_{(1)}(x) &= sqrt{1 - x^2} + x arcsin x \
F^{-1}_{(2)}(x) &= tfrac14 (3x sqrt{1 - x^2} + 2 x^2 arcsin x + arcsin x) \
F^{-1}_{(1)}(f(x)) &= cos x + xsin x \
F^{-1}_{(2)}(f(x)) &= tfrac14 (3sin x cos x + 2 xsin^2 x + x)
end{align}



begin{align}
int sin^2 x,dx
&= x sin^2 x - 2sin x (cos x + xsin x)
+ 2 cdot tfrac14 (3sin x cos x + 2 xsin^2 x + x) + C \
&= x sin^2 x - 2sin x cos x - 2xsin^2 x
+ tfrac32sin x cos x + xsin^2 x + tfrac12x + C \
&= (x - 2x +x)sin^2 x + left(- 2 + tfrac32right)sin x cos x + tfrac12x + C \
&= -tfrac12 sin x cos x + tfrac12x + C. \
end{align}





begin{align}
f(x) &= e^x \
f^{-1}(x) &= ln x \
F^{-1}_{(1)}(x) &= x (ln x - 1) \
F^{-1}_{(2)}(x) &= tfrac14 x^2 (2 ln x - 3) \
F^{-1}_{(1)}(f(x)) &= e^x (x - 1) \
F^{-1}_{(2)}(f(x)) &= tfrac14 e^{2x} (2 x - 3)
end{align}



begin{align}
int (e^x)^2,dx
&= x e^{2x} - 2e^x cdot e^x (x - 1) + 2 cdot tfrac14 e^{2x} (2 x - 3) + C \
&= x e^{2x} - 2(x - 1)e^{2x} + left(x - tfrac32right) e^{2x} + C \
&= left(x - 2x + 2 + x - tfrac32right)e^{2x} + C \
&= tfrac12 e^{2x} + C. \
end{align}





begin{align}
f(x) &= e^{x^2} \
f^{-1}(x) &= sqrt{ln x} \
F^{-1}_{(1)}(x) &= xsqrt{ln x} - tfrac12sqrtpi,mathrm{erfi}(sqrt{ln x}) \
F^{-1}_{(2)}(x)
&= -tfrac12 sqrtpi x , mathrm{erfi}(sqrt{ln x})
+ tfrac18sqrt{2pi},mathrm{erfi}(sqrt{2ln x}) + tfrac12 x^2 sqrt{ln x} \
x f(x)^2 &= x e^{2x^2}, \
F^{-1}_{(1)}(f(x)) &= x e^{x^2} - tfrac12sqrtpi,mathrm{erfi}(x) \
2f(x)F^{-1}_{(1)}(f(x)) &= 2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x),\
2F^{-1}_{(2)}(f(x)) &= -sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} \
end{align}



begin{align}
int left(e^{x^2}right)^2,dx
&= x e^{2x^2} - left(2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x)right)
- sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} + C\
&= tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + C.
end{align}



This agrees with $int e^{u^2},du = tfrac12sqrtpi,mathrm{erfi}(u) + C$
with the substitution $u = sqrt2 x.$






share|cite|improve this answer












My first thought was that if this were true, I'd have seen it already in conjunction with the tables of integrals in one of the standard reference books.
But it seems it is true.



The reason the formula isn't more prominent may be because it requires $f(x)$ to have an inverse, and then it requires that inverse to be integrated twice.
Those requirements put a lot of functions out of reach of this formula.
For example, consider $f(x) = xe^{x^2/2},$ which is sometimes given as an example of why there (supposedly) is no general formula to integrate $f(x).$
(For example, see
how to calculate integral of square of a function
and Integral of Derivative squared).
Then $f^{-1}(x) = sqrt{mathrm W(x^2)},$ where W is the Lambert W function, but
(according to Wolfram Alpha) $sqrt{mathrm W(x^2)}$ has no integral expressible in elementary functions.



Out of curiosity, I worked this out for a few examples.
All of these are confirmed by other methods (which in these examples are
easier than the formula).





begin{align}
f(x) &= x \
f^{-1}(x) &= x \
F^{-1}_{(1)}(x) &= tfrac12 x^2 \
F^{-1}_{(2)}(x) &= tfrac16 x^3 \
F^{-1}_{(1)}(f(x)) &= tfrac12 x^2 \
F^{-1}_{(2)}(f(x)) &= tfrac16 x^3
end{align}



begin{align}
int x^2,dx
&= x cdot x^2 - 2x cdot tfrac12 x^2 + 2 cdot tfrac16 x^3 + C \
&= x^3 - x^3 + tfrac13 x^3 + C \
&= tfrac13 x^3 + C.
end{align}





begin{align}
f(x) &= sin x \
f^{-1}(x) &= arcsin x \
F^{-1}_{(1)}(x) &= sqrt{1 - x^2} + x arcsin x \
F^{-1}_{(2)}(x) &= tfrac14 (3x sqrt{1 - x^2} + 2 x^2 arcsin x + arcsin x) \
F^{-1}_{(1)}(f(x)) &= cos x + xsin x \
F^{-1}_{(2)}(f(x)) &= tfrac14 (3sin x cos x + 2 xsin^2 x + x)
end{align}



begin{align}
int sin^2 x,dx
&= x sin^2 x - 2sin x (cos x + xsin x)
+ 2 cdot tfrac14 (3sin x cos x + 2 xsin^2 x + x) + C \
&= x sin^2 x - 2sin x cos x - 2xsin^2 x
+ tfrac32sin x cos x + xsin^2 x + tfrac12x + C \
&= (x - 2x +x)sin^2 x + left(- 2 + tfrac32right)sin x cos x + tfrac12x + C \
&= -tfrac12 sin x cos x + tfrac12x + C. \
end{align}





begin{align}
f(x) &= e^x \
f^{-1}(x) &= ln x \
F^{-1}_{(1)}(x) &= x (ln x - 1) \
F^{-1}_{(2)}(x) &= tfrac14 x^2 (2 ln x - 3) \
F^{-1}_{(1)}(f(x)) &= e^x (x - 1) \
F^{-1}_{(2)}(f(x)) &= tfrac14 e^{2x} (2 x - 3)
end{align}



begin{align}
int (e^x)^2,dx
&= x e^{2x} - 2e^x cdot e^x (x - 1) + 2 cdot tfrac14 e^{2x} (2 x - 3) + C \
&= x e^{2x} - 2(x - 1)e^{2x} + left(x - tfrac32right) e^{2x} + C \
&= left(x - 2x + 2 + x - tfrac32right)e^{2x} + C \
&= tfrac12 e^{2x} + C. \
end{align}





begin{align}
f(x) &= e^{x^2} \
f^{-1}(x) &= sqrt{ln x} \
F^{-1}_{(1)}(x) &= xsqrt{ln x} - tfrac12sqrtpi,mathrm{erfi}(sqrt{ln x}) \
F^{-1}_{(2)}(x)
&= -tfrac12 sqrtpi x , mathrm{erfi}(sqrt{ln x})
+ tfrac18sqrt{2pi},mathrm{erfi}(sqrt{2ln x}) + tfrac12 x^2 sqrt{ln x} \
x f(x)^2 &= x e^{2x^2}, \
F^{-1}_{(1)}(f(x)) &= x e^{x^2} - tfrac12sqrtpi,mathrm{erfi}(x) \
2f(x)F^{-1}_{(1)}(f(x)) &= 2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x),\
2F^{-1}_{(2)}(f(x)) &= -sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} \
end{align}



begin{align}
int left(e^{x^2}right)^2,dx
&= x e^{2x^2} - left(2x e^{2x^2} - sqrtpi,e^{x^2}mathrm{erfi}(x)right)
- sqrtpi e^{x^2} mathrm{erfi}(x)
+ tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + xe^{2x^2} + C\
&= tfrac14sqrt{2pi}, mathrm{erfi}(sqrt2 x) + C.
end{align}



This agrees with $int e^{u^2},du = tfrac12sqrtpi,mathrm{erfi}(u) + C$
with the substitution $u = sqrt2 x.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 16:22









David K

52.6k340115




52.6k340115








  • 1




    Thank you for that analysis. So it seems that this formula works nicely only for certain functions, for others it is unnecessarily complicating. It seems that the best functions to integrate this method are powers of logarithms
    – KKZiomek
    Dec 9 '18 at 18:04






  • 1




    It did work out very nicely for $(ln x)^2$; no cancellation of unnecessary terms. I would not be surprised if there were some other functions it worked nicely for.
    – David K
    Dec 10 '18 at 0:35






  • 1




    True, but arguably $ln x $ is the *nicest* of all, as its inverse is $e^x$ whose antiderivtives are all $e^x$. Though, like you said, there may be functions that are nice enough for this method besides logarithms
    – KKZiomek
    Dec 10 '18 at 5:08














  • 1




    Thank you for that analysis. So it seems that this formula works nicely only for certain functions, for others it is unnecessarily complicating. It seems that the best functions to integrate this method are powers of logarithms
    – KKZiomek
    Dec 9 '18 at 18:04






  • 1




    It did work out very nicely for $(ln x)^2$; no cancellation of unnecessary terms. I would not be surprised if there were some other functions it worked nicely for.
    – David K
    Dec 10 '18 at 0:35






  • 1




    True, but arguably $ln x $ is the *nicest* of all, as its inverse is $e^x$ whose antiderivtives are all $e^x$. Though, like you said, there may be functions that are nice enough for this method besides logarithms
    – KKZiomek
    Dec 10 '18 at 5:08








1




1




Thank you for that analysis. So it seems that this formula works nicely only for certain functions, for others it is unnecessarily complicating. It seems that the best functions to integrate this method are powers of logarithms
– KKZiomek
Dec 9 '18 at 18:04




Thank you for that analysis. So it seems that this formula works nicely only for certain functions, for others it is unnecessarily complicating. It seems that the best functions to integrate this method are powers of logarithms
– KKZiomek
Dec 9 '18 at 18:04




1




1




It did work out very nicely for $(ln x)^2$; no cancellation of unnecessary terms. I would not be surprised if there were some other functions it worked nicely for.
– David K
Dec 10 '18 at 0:35




It did work out very nicely for $(ln x)^2$; no cancellation of unnecessary terms. I would not be surprised if there were some other functions it worked nicely for.
– David K
Dec 10 '18 at 0:35




1




1




True, but arguably $ln x $ is the *nicest* of all, as its inverse is $e^x$ whose antiderivtives are all $e^x$. Though, like you said, there may be functions that are nice enough for this method besides logarithms
– KKZiomek
Dec 10 '18 at 5:08




True, but arguably $ln x $ is the *nicest* of all, as its inverse is $e^x$ whose antiderivtives are all $e^x$. Though, like you said, there may be functions that are nice enough for this method besides logarithms
– KKZiomek
Dec 10 '18 at 5:08


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032003%2fdid-i-just-discover-this-integration-formula%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Plaza Victoria

Puebla de Zaragoza

Musa