1. ## 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

$
\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

2. 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.

3. $
\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 $dv=1$

4. 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.

5. 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

6. 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