Evaluate $int frac{sqrt{64x^2-256}}{x},dx$












0












$begingroup$



$$int frac{sqrt{64x^2-256}}{x},dx$$




Image.



I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it



enter image description here




$newcommand{dd}{; mathrm{d}}int frac{sqrt{64x^2-256}}x dd x to
int frac{sqrt{64(x^2-4)}}x dd x to
int frac{8sqrt{x^2-4}}x dd x$



Use $x=asectheta$, $dd x=asectheta tantheta dd theta$.



$a=2$ $to$ $x=2sectheta$, $dd x=2sectheta tantheta dd theta$.



$=int frac{8sqrt{4sec^2theta-4}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8sqrt{4(sec^2theta-1)}}{2sectheta}(2secthetatantheta) dd theta $



$=int frac{8sqrt{4tan^2theta}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8(2tantheta)}{2sectheta}(2secthetatantheta) dd theta$



$=int 16tan^2theta dd theta to
16inttan^2theta dd theta to
underset{text{trig. formula}}{underbrace{16(theta+tantheta)+C}}$



$Rightarrow 16(tantheta-theta)+C = 16tantheta-16theta$



$x=2sectheta$, $sectheta= frac x2$



$boxed{16tanleft(frac{sqrt{x^2-4}}2right) -16sec^{-1}left(frac x2right)+C}$







calculus integration indefinite-integrals radicals







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  • $begingroup$
    Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
    $endgroup$
    – Em.
    Jul 13 '16 at 4:07






  • 3




    $begingroup$
    The substitution $x=frac{2}{sintheta}$ gives an easy integral.
    $endgroup$
    – Jack D'Aurizio
    Jul 13 '16 at 4:09










  • $begingroup$
    I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
    $endgroup$
    – Martin Sleziak
    Jul 27 '16 at 8:51
















0












$begingroup$



$$int frac{sqrt{64x^2-256}}{x},dx$$




Image.



I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it



enter image description here




$newcommand{dd}{; mathrm{d}}int frac{sqrt{64x^2-256}}x dd x to
int frac{sqrt{64(x^2-4)}}x dd x to
int frac{8sqrt{x^2-4}}x dd x$



Use $x=asectheta$, $dd x=asectheta tantheta dd theta$.



$a=2$ $to$ $x=2sectheta$, $dd x=2sectheta tantheta dd theta$.



$=int frac{8sqrt{4sec^2theta-4}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8sqrt{4(sec^2theta-1)}}{2sectheta}(2secthetatantheta) dd theta $



$=int frac{8sqrt{4tan^2theta}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8(2tantheta)}{2sectheta}(2secthetatantheta) dd theta$



$=int 16tan^2theta dd theta to
16inttan^2theta dd theta to
underset{text{trig. formula}}{underbrace{16(theta+tantheta)+C}}$



$Rightarrow 16(tantheta-theta)+C = 16tantheta-16theta$



$x=2sectheta$, $sectheta= frac x2$



$boxed{16tanleft(frac{sqrt{x^2-4}}2right) -16sec^{-1}left(frac x2right)+C}$







calculus integration indefinite-integrals radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
    $endgroup$
    – Em.
    Jul 13 '16 at 4:07






  • 3




    $begingroup$
    The substitution $x=frac{2}{sintheta}$ gives an easy integral.
    $endgroup$
    – Jack D'Aurizio
    Jul 13 '16 at 4:09










  • $begingroup$
    I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
    $endgroup$
    – Martin Sleziak
    Jul 27 '16 at 8:51














0












0








0





$begingroup$



$$int frac{sqrt{64x^2-256}}{x},dx$$




Image.



I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it



enter image description here




$newcommand{dd}{; mathrm{d}}int frac{sqrt{64x^2-256}}x dd x to
int frac{sqrt{64(x^2-4)}}x dd x to
int frac{8sqrt{x^2-4}}x dd x$



Use $x=asectheta$, $dd x=asectheta tantheta dd theta$.



$a=2$ $to$ $x=2sectheta$, $dd x=2sectheta tantheta dd theta$.



$=int frac{8sqrt{4sec^2theta-4}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8sqrt{4(sec^2theta-1)}}{2sectheta}(2secthetatantheta) dd theta $



$=int frac{8sqrt{4tan^2theta}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8(2tantheta)}{2sectheta}(2secthetatantheta) dd theta$



$=int 16tan^2theta dd theta to
16inttan^2theta dd theta to
underset{text{trig. formula}}{underbrace{16(theta+tantheta)+C}}$



$Rightarrow 16(tantheta-theta)+C = 16tantheta-16theta$



$x=2sectheta$, $sectheta= frac x2$



$boxed{16tanleft(frac{sqrt{x^2-4}}2right) -16sec^{-1}left(frac x2right)+C}$







calculus integration indefinite-integrals radicals







share|cite|improve this question











$endgroup$





$$int frac{sqrt{64x^2-256}}{x},dx$$




Image.



I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it



enter image description here




$newcommand{dd}{; mathrm{d}}int frac{sqrt{64x^2-256}}x dd x to
int frac{sqrt{64(x^2-4)}}x dd x to
int frac{8sqrt{x^2-4}}x dd x$



Use $x=asectheta$, $dd x=asectheta tantheta dd theta$.



$a=2$ $to$ $x=2sectheta$, $dd x=2sectheta tantheta dd theta$.



$=int frac{8sqrt{4sec^2theta-4}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8sqrt{4(sec^2theta-1)}}{2sectheta}(2secthetatantheta) dd theta $



$=int frac{8sqrt{4tan^2theta}}{2sectheta}(2secthetatantheta) dd theta to
int frac{8(2tantheta)}{2sectheta}(2secthetatantheta) dd theta$



$=int 16tan^2theta dd theta to
16inttan^2theta dd theta to
underset{text{trig. formula}}{underbrace{16(theta+tantheta)+C}}$



$Rightarrow 16(tantheta-theta)+C = 16tantheta-16theta$



$x=2sectheta$, $sectheta= frac x2$



$boxed{16tanleft(frac{sqrt{x^2-4}}2right) -16sec^{-1}left(frac x2right)+C}$







calculus integration indefinite-integrals radicals




calculus integration indefinite-integrals radicals






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share|cite|improve this question













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edited Jul 27 '16 at 8:50









Martin Sleziak

44.7k9117272




44.7k9117272










asked Jul 13 '16 at 4:03









cchangcchang

11




11












  • $begingroup$
    Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
    $endgroup$
    – Em.
    Jul 13 '16 at 4:07






  • 3




    $begingroup$
    The substitution $x=frac{2}{sintheta}$ gives an easy integral.
    $endgroup$
    – Jack D'Aurizio
    Jul 13 '16 at 4:09










  • $begingroup$
    I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
    $endgroup$
    – Martin Sleziak
    Jul 27 '16 at 8:51


















  • $begingroup$
    Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
    $endgroup$
    – Em.
    Jul 13 '16 at 4:07






  • 3




    $begingroup$
    The substitution $x=frac{2}{sintheta}$ gives an easy integral.
    $endgroup$
    – Jack D'Aurizio
    Jul 13 '16 at 4:09










  • $begingroup$
    I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
    $endgroup$
    – Martin Sleziak
    Jul 27 '16 at 8:51
















$begingroup$
Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
$endgroup$
– Em.
Jul 13 '16 at 4:07




$begingroup$
Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here.
$endgroup$
– Em.
Jul 13 '16 at 4:07




3




3




$begingroup$
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
$endgroup$
– Jack D'Aurizio
Jul 13 '16 at 4:09




$begingroup$
The substitution $x=frac{2}{sintheta}$ gives an easy integral.
$endgroup$
– Jack D'Aurizio
Jul 13 '16 at 4:09












$begingroup$
I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
$endgroup$
– Martin Sleziak
Jul 27 '16 at 8:51




$begingroup$
I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say.
$endgroup$
– Martin Sleziak
Jul 27 '16 at 8:51










3 Answers
3






active

oldest

votes


















5












$begingroup$

Since I am almost blind, I have a lot of problems reading the image.



Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
    $endgroup$
    – cchang
    Jul 13 '16 at 4:38










  • $begingroup$
    If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
    $endgroup$
    – Claude Leibovici
    Jul 13 '16 at 4:45



















1












$begingroup$

$$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$



Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$



$$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @user376343,Thanks for your feedback
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 17:07



















0












$begingroup$

$newcommand{angles}[1]{leftlangle,{#1},rightrangle}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{half}{{1 over 2}}
newcommand{ic}{mathrm{i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{Li}[1]{,mathrm{Li}}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$




With the
sub$ds{ldots t equiv x - root{x^{2} - 4} imp
x = {t^{2} + 4 over 2t}}$:






begin{align}
&color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
8int{root{x^{2} - 4} over x},dd x =
8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
\[3mm] = &
32arctanpars{t over 2} + {128 over t} + 24t
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
32arctanpars{x - root{x^{2} - 4} over 2} +
32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
\[3mm] = &
color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
end{align}




share|cite|improve this answer









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    3 Answers
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    active

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    3 Answers
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    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    5












    $begingroup$

    Since I am almost blind, I have a lot of problems reading the image.



    Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
      $endgroup$
      – cchang
      Jul 13 '16 at 4:38










    • $begingroup$
      If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
      $endgroup$
      – Claude Leibovici
      Jul 13 '16 at 4:45
















    5












    $begingroup$

    Since I am almost blind, I have a lot of problems reading the image.



    Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
      $endgroup$
      – cchang
      Jul 13 '16 at 4:38










    • $begingroup$
      If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
      $endgroup$
      – Claude Leibovici
      Jul 13 '16 at 4:45














    5












    5








    5





    $begingroup$

    Since I am almost blind, I have a lot of problems reading the image.



    Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$






    share|cite|improve this answer









    $endgroup$



    Since I am almost blind, I have a lot of problems reading the image.



    Consider $$I=int frac{sqrt{64 x^2-256}}{x},dx$$ What you apparently did is $x=2sec(t)$, $dx=2 tan (t) sec (t)$ which make $$I=int tan (t) sqrt{256 sec ^2(t)-256},dt=16int tan (t) sqrt{tan ^2(t)},dt=16int tan^2 (t) ,dt$$ $$I=16int (1+tan^2(t)-1),dt=16 (tan (t)-t)$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jul 13 '16 at 4:25









    Claude LeiboviciClaude Leibovici

    120k1157132




    120k1157132












    • $begingroup$
      Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
      $endgroup$
      – cchang
      Jul 13 '16 at 4:38










    • $begingroup$
      If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
      $endgroup$
      – Claude Leibovici
      Jul 13 '16 at 4:45


















    • $begingroup$
      Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
      $endgroup$
      – cchang
      Jul 13 '16 at 4:38










    • $begingroup$
      If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
      $endgroup$
      – Claude Leibovici
      Jul 13 '16 at 4:45
















    $begingroup$
    Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
    $endgroup$
    – cchang
    Jul 13 '16 at 4:38




    $begingroup$
    Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct.
    $endgroup$
    – cchang
    Jul 13 '16 at 4:38












    $begingroup$
    If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
    $endgroup$
    – Claude Leibovici
    Jul 13 '16 at 4:45




    $begingroup$
    If you properly replace $t$, you should arrive to $2 sqrt{x^2-64}+16 tan ^{-1}left(frac{8}{sqrt{x^2-64}}right)$
    $endgroup$
    – Claude Leibovici
    Jul 13 '16 at 4:45











    1












    $begingroup$

    $$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$



    Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$



    $$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @user376343,Thanks for your feedback
      $endgroup$
      – lab bhattacharjee
      Nov 30 '18 at 17:07
















    1












    $begingroup$

    $$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$



    Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$



    $$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @user376343,Thanks for your feedback
      $endgroup$
      – lab bhattacharjee
      Nov 30 '18 at 17:07














    1












    1








    1





    $begingroup$

    $$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$



    Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$



    $$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$






    share|cite|improve this answer











    $endgroup$



    $$dfrac{sqrt{64x^2-256}}x=8xcdotdfrac{sqrt{x^2-4}}{x^2}$$



    Let $sqrt{x^2-4}=yimplies x^2-4=y^2implies x dx= y dy$



    $$intdfrac{sqrt{64x^2-256}}xdx=8intdfrac{y^2dy}{y^2+4}=8int dy-32intdfrac{dy}{y^2+4}=?$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 30 '18 at 17:06

























    answered Jul 13 '16 at 6:05









    lab bhattacharjeelab bhattacharjee

    224k15156274




    224k15156274












    • $begingroup$
      @user376343,Thanks for your feedback
      $endgroup$
      – lab bhattacharjee
      Nov 30 '18 at 17:07


















    • $begingroup$
      @user376343,Thanks for your feedback
      $endgroup$
      – lab bhattacharjee
      Nov 30 '18 at 17:07
















    $begingroup$
    @user376343,Thanks for your feedback
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 17:07




    $begingroup$
    @user376343,Thanks for your feedback
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 17:07











    0












    $begingroup$

    $newcommand{angles}[1]{leftlangle,{#1},rightrangle}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
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    newcommand{ol}[1]{overline{#1}}
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    newcommand{ul}[1]{underline{#1}}
    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$




    With the
    sub$ds{ldots t equiv x - root{x^{2} - 4} imp
    x = {t^{2} + 4 over 2t}}$:






    begin{align}
    &color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
    8int{root{x^{2} - 4} over x},dd x =
    8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
    \[3mm] = &
    32arctanpars{t over 2} + {128 over t} + 24t
    \[3mm] = &
    32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
    \[3mm] = &
    32arctanpars{x - root{x^{2} - 4} over 2} +
    32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
    \[3mm] = &
    color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
    56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
    end{align}




    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $newcommand{angles}[1]{leftlangle,{#1},rightrangle}
      newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
      newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
      newcommand{dd}{mathrm{d}}
      newcommand{ds}[1]{displaystyle{#1}}
      newcommand{expo}[1]{,mathrm{e}^{#1},}
      newcommand{half}{{1 over 2}}
      newcommand{ic}{mathrm{i}}
      newcommand{iff}{Longleftrightarrow}
      newcommand{imp}{Longrightarrow}
      newcommand{Li}[1]{,mathrm{Li}}
      newcommand{ol}[1]{overline{#1}}
      newcommand{pars}[1]{left(,{#1},right)}
      newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{ul}[1]{underline{#1}}
      newcommand{root}[2]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$




      With the
      sub$ds{ldots t equiv x - root{x^{2} - 4} imp
      x = {t^{2} + 4 over 2t}}$:






      begin{align}
      &color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
      8int{root{x^{2} - 4} over x},dd x =
      8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
      \[3mm] = &
      32arctanpars{t over 2} + {128 over t} + 24t
      \[3mm] = &
      32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
      \[3mm] = &
      32arctanpars{x - root{x^{2} - 4} over 2} +
      32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
      \[3mm] = &
      color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
      56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
      end{align}




      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $newcommand{angles}[1]{leftlangle,{#1},rightrangle}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{half}{{1 over 2}}
        newcommand{ic}{mathrm{i}}
        newcommand{iff}{Longleftrightarrow}
        newcommand{imp}{Longrightarrow}
        newcommand{Li}[1]{,mathrm{Li}}
        newcommand{ol}[1]{overline{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{ul}[1]{underline{#1}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$




        With the
        sub$ds{ldots t equiv x - root{x^{2} - 4} imp
        x = {t^{2} + 4 over 2t}}$:






        begin{align}
        &color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
        8int{root{x^{2} - 4} over x},dd x =
        8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
        \[3mm] = &
        32arctanpars{t over 2} + {128 over t} + 24t
        \[3mm] = &
        32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
        \[3mm] = &
        32arctanpars{x - root{x^{2} - 4} over 2} +
        32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
        \[3mm] = &
        color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
        56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
        end{align}




        share|cite|improve this answer









        $endgroup$



        $newcommand{angles}[1]{leftlangle,{#1},rightrangle}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{half}{{1 over 2}}
        newcommand{ic}{mathrm{i}}
        newcommand{iff}{Longleftrightarrow}
        newcommand{imp}{Longrightarrow}
        newcommand{Li}[1]{,mathrm{Li}}
        newcommand{ol}[1]{overline{#1}}
        newcommand{pars}[1]{left(,{#1},right)}
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{ul}[1]{underline{#1}}
        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$




        With the
        sub$ds{ldots t equiv x - root{x^{2} - 4} imp
        x = {t^{2} + 4 over 2t}}$:






        begin{align}
        &color{#f00}{int{root{64x^{2} - 256} over x},dd x} =
        8int{root{x^{2} - 4} over x},dd x =
        8intpars{{8 over t^{2} + 4} - {16 over t^{2}} + 3},dd t
        \[3mm] = &
        32arctanpars{t over 2} + {128 over t} + 24t
        \[3mm] = &
        32arctanpars{x - root{x^{2} - 4} over 2} + {128 over x - root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
        \[3mm] = &
        32arctanpars{x - root{x^{2} - 4} over 2} +
        32pars{x + root{x^{2} - 4}} + 24pars{x - root{x^{2} - 4}}
        \[3mm] = &
        color{#f00}{32arctanpars{x - root{x^{2} - 4} over 2} +
        56x + 8root{x^{2} - 4}} + pars{~mbox{a constant}~}
        end{align}





        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jul 13 '16 at 5:41









        Felix MarinFelix Marin

        67.4k7107141




        67.4k7107141






























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