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Math Help - integration by parts

  1. #1
    Junior Member scottie.mcdonald's Avatar
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    integration by parts

    i apologize about my last thread, correcting it took longer than 15 minutes, and i am unable to contact the director due to me not having microsoft on my computer. here is the question i meant to post. again, sorry for the inconvenience and trouble.

    evaluate the integral

    <br />
\int_1^{sqrt3} \arctan({1/x}) dx<br />

    this is what i have so far. i'm not sure if i even chose the right parts.

    <br />
f\prime(x)= arctan dx,<br />
f(x)=1/(1+x^2),<br />
g(x)= 1/x,<br />
g\prime(x)= ln |x|,<br />
    <br />
\int_1^{sqrt3} g\prime(x)f(x)dx = f(x)g(x) |_1^{sqrt3} - \int f\prime(x)g(x)dx<br />
    <br />
\int_1^{sqrt3} \frac{1}{1+x^2} (ln |x|)dx = \frac{1}{1+x^2}|_1^{sqrt3} - \int \frac{1}{xarctandx}<br />

    after this i don't know what to do (and i'm not sure if i've even begun correctly). any ideas or suggestions would be helpful.

    thank you,

    Scott
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  2. #2
    MHF Contributor
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    If this happens again, you should report the post by clicking the triangle bordered in red at the top-right of every post and that will notify us of the problem. You'll avoid an infraction as well.
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  3. #3
    Member Nacho's Avatar
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    <br />
\begin{gathered}<br />
  \int {\arctan \left( {\frac{1}<br />
{x}} \right)dx}  = x\arctan \left( {\frac{1}<br />
{x}} \right) + \int {x \cdot \frac{1}<br />
{{1 + \left( {\frac{1}<br />
{x}} \right)^2 }}}  \cdot \frac{1}<br />
{{x^2 }}dx =  \hfill \\<br />
   = x\arctan \left( {\frac{1}<br />
{x}} \right) + \int {\frac{1}<br />
{{x^2  + 1}}dx}  = x\arctan \left( {\frac{1}<br />
{x}} \right) + \arctan \left( x \right) + C \hfill \\ <br />
\end{gathered} <br />

    Whereat u=\arctan \left(\dfrac{1}{x} \right) and dv=1
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  4. #4
    Junior Member scottie.mcdonald's Avatar
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    Quote Originally Posted by scottie.mcdonald View Post

    <br />
\int_1^{sqrt3} \frac{1}{1+x^2} (ln |x|)dx = \frac{1}{1+x^2}|_1^{sqrt3} - \int \frac{1}{xarctandx}<br />
    this is what i now have after some more thought:

    <br />
\int_1^{sqrt3} \frac{ln|x|dx}{1+x^2} = \frac{1}{3+3 \sqrt3}- \frac{1}{2} - \int_1^{sqrt3} \frac{1}{xarctandx}<br />

    again, i think i'm going in the wrong direction with the question, but i don't know where to go.
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  5. #5
    Junior Member scottie.mcdonald's Avatar
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    Quote Originally Posted by Nacho View Post
    <br />
\begin{gathered}<br />
\int {\arctan \left( {\frac{1}<br />
{x}} \right)dx} = x\arctan \left( {\frac{1}<br />
{x}} \right) + \int {x \cdot \frac{1}<br />
{{1 + \left( {\frac{1}<br />
{x}} \right)^2 }}} \cdot \frac{1}<br />
{{x^2 }}dx = \hfill \\<br />
= x\arctan \left( {\frac{1}<br />
{x}} \right) + \int {\frac{1}<br />
{{x^2 + 1}}dx} = x\arctan \left( {\frac{1}<br />
{x}} \right) + \arctan \left( x \right) + C \hfill \\ <br />
\end{gathered} <br />

    Whereat u=\arctan \left(\dfrac{1}{x} \right) and dv=1

    thank you Nacho, i was writing when you submitted that and didn't see it.

    Thanks for the help =)

    Scott
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  6. #6
    Member Nacho's Avatar
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    Quote Originally Posted by scottie.mcdonald View Post
    thank you Nacho, i was writing when you submitted that and didn't see it.

    Thanks for the help =)

    Scott
    you´re welcome, but I make a mistake

    <br />
x \cdot \frac{1}<br />
{{1 + \left( {\frac{1}<br />
{x}} \right)^2 }} \cdot \frac{1}<br />
{{x^2 }} = \frac{x}<br />
{{x^2  + 1}}<br />

    But this is easy integrate:

    <br />
\begin{gathered}<br />
  {\text{Knowing:}} \hfill \\<br />
  \int {\frac{{f'(x)}}<br />
{{f(x)}}dx = \ln \left| {f(x)} \right| + C}  \hfill \\<br />
   \hfill \\<br />
  {\text{Then: }}\int {\frac{x}<br />
{{x^2  + 1}}dx}  = \frac{1}<br />
{2}\int {\frac{{2x}}<br />
{{x^2  + 1}}dx}  = \frac{1}<br />
{2}\int {\frac{{\left( {x^2  + 1} \right)^\prime  }}<br />
{{x^2  + 1}}dx}  \hfill \\ <br />
\end{gathered} <br />

    Sorry
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