Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
up vote
1
down vote
favorite
Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
calculus real-analysis integration analysis
add a comment |
up vote
1
down vote
favorite
Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
calculus real-analysis integration analysis
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
calculus real-analysis integration analysis
Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
calculus real-analysis integration analysis
calculus real-analysis integration analysis
edited Nov 19 at 11:34
asked Nov 19 at 10:25
hopefully
132112
132112
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
Nov 19 at 11:07
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
Nov 19 at 11:19
No I do not mean this
– hopefully
Nov 19 at 11:25
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
Nov 19 at 11:26
I will include it in the main question above
– hopefully
Nov 19 at 11:27
|
show 2 more comments
up vote
2
down vote
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
why you see that it is incorrect?
– hopefully
Nov 20 at 5:15
@hopefully Click this link: desmos.com/calculator/hojmgfvkqb
– clathratus
Nov 20 at 5:20
I have clicked it ...... I think the problem in your first solution because you decompose as if there were difference between 2 squares which is incorrect.
– hopefully
Nov 20 at 5:22
So I think the solution in the book is wrong.
– hopefully
Nov 20 at 5:22
1
@hopefully the book isn't wrong, but I don't know why $$3^{-1/2}arctanbigg((3^{1/2}x)^{-1}(x^2+1)bigg)$$ is wrong
– clathratus
Nov 20 at 5:33
|
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
Nov 19 at 11:07
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
Nov 19 at 11:19
No I do not mean this
– hopefully
Nov 19 at 11:25
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
Nov 19 at 11:26
I will include it in the main question above
– hopefully
Nov 19 at 11:27
|
show 2 more comments
up vote
2
down vote
accepted
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
Nov 19 at 11:07
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
Nov 19 at 11:19
No I do not mean this
– hopefully
Nov 19 at 11:25
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
Nov 19 at 11:26
I will include it in the main question above
– hopefully
Nov 19 at 11:27
|
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
answered Nov 19 at 10:31
crskhr
3,760925
3,760925
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
Nov 19 at 11:07
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
Nov 19 at 11:19
No I do not mean this
– hopefully
Nov 19 at 11:25
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
Nov 19 at 11:26
I will include it in the main question above
– hopefully
Nov 19 at 11:27
|
show 2 more comments
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
Nov 19 at 11:07
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
Nov 19 at 11:19
No I do not mean this
– hopefully
Nov 19 at 11:25
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
Nov 19 at 11:26
I will include it in the main question above
– hopefully
Nov 19 at 11:27
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
Nov 19 at 11:07
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
Nov 19 at 11:07
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
Nov 19 at 11:19
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
Nov 19 at 11:19
No I do not mean this
– hopefully
Nov 19 at 11:25
No I do not mean this
– hopefully
Nov 19 at 11:25
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
Nov 19 at 11:26
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
Nov 19 at 11:26
I will include it in the main question above
– hopefully
Nov 19 at 11:27
I will include it in the main question above
– hopefully
Nov 19 at 11:27
|
show 2 more comments
up vote
2
down vote
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
why you see that it is incorrect?
– hopefully
Nov 20 at 5:15
@hopefully Click this link: desmos.com/calculator/hojmgfvkqb
– clathratus
Nov 20 at 5:20
I have clicked it ...... I think the problem in your first solution because you decompose as if there were difference between 2 squares which is incorrect.
– hopefully
Nov 20 at 5:22
So I think the solution in the book is wrong.
– hopefully
Nov 20 at 5:22
1
@hopefully the book isn't wrong, but I don't know why $$3^{-1/2}arctanbigg((3^{1/2}x)^{-1}(x^2+1)bigg)$$ is wrong
– clathratus
Nov 20 at 5:33
|
show 1 more comment
up vote
2
down vote
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
why you see that it is incorrect?
– hopefully
Nov 20 at 5:15
@hopefully Click this link: desmos.com/calculator/hojmgfvkqb
– clathratus
Nov 20 at 5:20
I have clicked it ...... I think the problem in your first solution because you decompose as if there were difference between 2 squares which is incorrect.
– hopefully
Nov 20 at 5:22
So I think the solution in the book is wrong.
– hopefully
Nov 20 at 5:22
1
@hopefully the book isn't wrong, but I don't know why $$3^{-1/2}arctanbigg((3^{1/2}x)^{-1}(x^2+1)bigg)$$ is wrong
– clathratus
Nov 20 at 5:33
|
show 1 more comment
up vote
2
down vote
up vote
2
down vote
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
answered Nov 20 at 0:11
clathratus
2,377324
2,377324
why you see that it is incorrect?
– hopefully
Nov 20 at 5:15
@hopefully Click this link: desmos.com/calculator/hojmgfvkqb
– clathratus
Nov 20 at 5:20
I have clicked it ...... I think the problem in your first solution because you decompose as if there were difference between 2 squares which is incorrect.
– hopefully
Nov 20 at 5:22
So I think the solution in the book is wrong.
– hopefully
Nov 20 at 5:22
1
@hopefully the book isn't wrong, but I don't know why $$3^{-1/2}arctanbigg((3^{1/2}x)^{-1}(x^2+1)bigg)$$ is wrong
– clathratus
Nov 20 at 5:33
|
show 1 more comment
why you see that it is incorrect?
– hopefully
Nov 20 at 5:15
@hopefully Click this link: desmos.com/calculator/hojmgfvkqb
– clathratus
Nov 20 at 5:20
I have clicked it ...... I think the problem in your first solution because you decompose as if there were difference between 2 squares which is incorrect.
– hopefully
Nov 20 at 5:22
So I think the solution in the book is wrong.
– hopefully
Nov 20 at 5:22
1
@hopefully the book isn't wrong, but I don't know why $$3^{-1/2}arctanbigg((3^{1/2}x)^{-1}(x^2+1)bigg)$$ is wrong
– clathratus
Nov 20 at 5:33
why you see that it is incorrect?
– hopefully
Nov 20 at 5:15
why you see that it is incorrect?
– hopefully
Nov 20 at 5:15
@hopefully Click this link: desmos.com/calculator/hojmgfvkqb
– clathratus
Nov 20 at 5:20
@hopefully Click this link: desmos.com/calculator/hojmgfvkqb
– clathratus
Nov 20 at 5:20
I have clicked it ...... I think the problem in your first solution because you decompose as if there were difference between 2 squares which is incorrect.
– hopefully
Nov 20 at 5:22
I have clicked it ...... I think the problem in your first solution because you decompose as if there were difference between 2 squares which is incorrect.
– hopefully
Nov 20 at 5:22
So I think the solution in the book is wrong.
– hopefully
Nov 20 at 5:22
So I think the solution in the book is wrong.
– hopefully
Nov 20 at 5:22
1
1
@hopefully the book isn't wrong, but I don't know why $$3^{-1/2}arctanbigg((3^{1/2}x)^{-1}(x^2+1)bigg)$$ is wrong
– clathratus
Nov 20 at 5:33
@hopefully the book isn't wrong, but I don't know why $$3^{-1/2}arctanbigg((3^{1/2}x)^{-1}(x^2+1)bigg)$$ is wrong
– clathratus
Nov 20 at 5:33
|
show 1 more comment
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