I'm having a hard time doing this problem.

(Tan^-1(x)) / 1 + x^2

I'm know i'm gonna have to use the substitution rule

I designated

u = arctan

du = 1 / (1 + x^2) dx

From here i'm fairly stuck. Any help is appreciated

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- Apr 24th 2008, 12:34 PMJonathanEyoonIntegration Problem
I'm having a hard time doing this problem.

(Tan^-1(x)) / 1 + x^2

I'm know i'm gonna have to use the substitution rule

I designated

u = arctan

du = 1 / (1 + x^2) dx

From here i'm fairly stuck. Any help is appreciated - Apr 24th 2008, 12:35 PMMathstud28
- Apr 24th 2008, 12:36 PMMathstud28
- Apr 24th 2008, 12:38 PMMoo
Hello,

You're right for the beginning ! :)

$\displaystyle u=\arctan(x)$

$\displaystyle du=\frac{1}{1+x^2} dx \rightarrow dx=(1+x^2)du$

Then $\displaystyle \int \frac{\arctan(x)}{1+x^2} dx=\int \frac{u}{1+x^2} dx=\int \frac{u}{\color{red}1+x^2} ({\color{red}1+x^2}) du$$\displaystyle =\int udu$ - Apr 24th 2008, 12:39 PMo_O
??? You're stuck? You just substitute it in ...

$\displaystyle {\color{red} u = \arctan x}$

$\displaystyle {\color{blue} du = \frac{1}{1 +x^{2}} dx}$

Looking at your integral:

$\displaystyle \int \frac{{\color{red}\arctan x}}{{\color{blue}1 + x^{2}}} {\color{blue} dx} = \int \left({\color{red}\arctan x} \cdot {\color{blue}\frac{1}{1+x^{2}}dx}\right) = \int {\color{red} u} {\color{blue}du}$ - Apr 24th 2008, 12:40 PMMathstud28
- Apr 24th 2008, 12:41 PMMoo
At least we explained why :D

We already discussed about that, it's no competition :) - Apr 24th 2008, 12:41 PMo_O
(Tongueout) Next time stud ...

Edit: omg moo beat me again. - Apr 24th 2008, 12:44 PMMathstud28
- Apr 24th 2008, 01:20 PMJonathanEyoon
Thanks guys!