integration by parts

**scottie.mcdonald** · February 26th 2009, 01:33 PM

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

$ \int_1^{sqrt3} \arctan({1/x}) dx $

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

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

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

**Jameson** · February 26th 2009, 01:54 PM

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.

**Nacho** · February 26th 2009, 01:54 PM

$ \begin{gathered} \int {\arctan \left( {\frac{1} {x}} \right)dx} = x\arctan \left( {\frac{1} {x}} \right) + \int {x \cdot \frac{1} {{1 + \left( {\frac{1} {x}} \right)^2 }}} \cdot \frac{1} {{x^2 }}dx = \hfill \\ = x\arctan \left( {\frac{1} {x}} \right) + \int {\frac{1} {{x^2 + 1}}dx} = x\arctan \left( {\frac{1} {x}} \right) + \arctan \left( x \right) + C \hfill \\ \end{gathered} $

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

**scottie.mcdonald** · February 26th 2009, 01:59 PM

Originally Posted by scottie.mcdonald

$ \int_1^{sqrt3} \frac{1}{1+x^2} (ln |x|)dx = \frac{1}{1+x^2}|_1^{sqrt3} - \int \frac{1}{xarctandx} $

this is what i now have after some more thought:

$ \int_1^{sqrt3} \frac{ln|x|dx}{1+x^2} = \frac{1}{3+3 \sqrt3}- \frac{1}{2} - \int_1^{sqrt3} \frac{1}{xarctandx} $

again, i think i'm going in the wrong direction with the question, but i don't know where to go.

**scottie.mcdonald** · February 26th 2009, 02:03 PM

Originally Posted by Nacho

$ \begin{gathered} \int {\arctan \left( {\frac{1} {x}} \right)dx} = x\arctan \left( {\frac{1} {x}} \right) + \int {x \cdot \frac{1} {{1 + \left( {\frac{1} {x}} \right)^2 }}} \cdot \frac{1} {{x^2 }}dx = \hfill \\ = x\arctan \left( {\frac{1} {x}} \right) + \int {\frac{1} {{x^2 + 1}}dx} = x\arctan \left( {\frac{1} {x}} \right) + \arctan \left( x \right) + C \hfill \\ \end{gathered} $

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

**Nacho** · February 26th 2009, 02:10 PM

Originally Posted by scottie.mcdonald

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

$ x \cdot \frac{1} {{1 + \left( {\frac{1} {x}} \right)^2 }} \cdot \frac{1} {{x^2 }} = \frac{x} {{x^2 + 1}} $

But this is easy integrate:

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

Sorry

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