Integration Problem

• Apr 24th 2008, 12:34 PM
JonathanEyoon
Integration 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 PM
Mathstud28
Quote:

Originally Posted by JonathanEyoon
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

You have the derivative of the quantity

therefore $\displaystyle \int\frac{arctan(x)}{1+x^2}=\frac{arctan^2(x)}{2}+ C$
• Apr 24th 2008, 12:36 PM
Mathstud28
Quote:

Originally Posted by JonathanEyoon
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

O and if you do it your way...so if you set it up $\displaystyle \int{u}du=\frac{u^2}{2}+C$

and remember $\displaystyle u=arctan(x)$
• Apr 24th 2008, 12:38 PM
Moo
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$$\displaystyle =\int udu • Apr 24th 2008, 12:39 PM o_O ??? You're stuck? You just substitute it in ... \displaystyle {\color{red} u = \arctan x} \displaystyle {\color{blue} du = \frac{1}{1 +x^{2}} dx} Looking at your integral: \displaystyle \int \frac{{\color{red}\arctan x}}{{\color{blue}1 + x^{2}}} {\color{blue} dx} = \int \left({\color{red}\arctan x} \cdot {\color{blue}\frac{1}{1+x^{2}}dx}\right) = \int {\color{red} u} {\color{blue}du} • Apr 24th 2008, 12:40 PM Mathstud28 Quote: Originally Posted by Moo Hello, You're right for the beginning ! :) \displaystyle u=\Arctan(x) \displaystyle du=\frac{1}{1+x^2} dx \rightarrow dx=(1+x^2)du Then \displaystyle \int \frac{\Arctan(x)}{1+x^2} dx=\int \frac{u}{1+x^2} dx=\int \frac{u}{\color{red}1+x^2} ({\color{red}1+x^2}) du$$\displaystyle =\int udu$

Quote:

Originally Posted by o_O
??? You're stuck? You just substitute it in ...

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

$\displaystyle \int \frac{{\color{red}\arctan x}}{{\color{blue}1 + x^{2}}} {\color{blue} dx} = \int \left({\color{red}\arctan x} \cdot {\color{blue}\frac{1}{1+x^{2}}dx}\right) = \int {\color{red} u} {\color{blue}du}$

Haha...too quick for you all (Cool)....just kidding (Rofl)
• Apr 24th 2008, 12:41 PM
Moo
At least we explained why :D
• Apr 24th 2008, 12:41 PM
o_O
(Tongueout) Next time stud ...

Edit: omg moo beat me again.
• Apr 24th 2008, 12:44 PM
Mathstud28
Quote:

Originally Posted by Moo
At least we explained why :D
I explained (Crying)...I said he had $\displaystyle \int{f(g(x))\cdot{g'(x)}}dx$