1. ## 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

2. 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 $\int\frac{arctan(x)}{1+x^2}=\frac{arctan^2(x)}{2}+ C$

3. 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 $\int{u}du=\frac{u^2}{2}+C$

and remember $u=arctan(x)$

4. Hello,

You're right for the beginning !

$u=\arctan(x)$

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

Then $\int \frac{\arctan(x)}{1+x^2} dx=\int \frac{u}{1+x^2} dx=\int \frac{u}{\color{red}1+x^2} ({\color{red}1+x^2}) du$ $=\int udu$

5. ??? You're stuck? You just substitute it in ...

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

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

6. Originally Posted by Moo
Hello,

You're right for the beginning !

$u=\Arctan(x)$

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

Then $\int \frac{\Arctan(x)}{1+x^2} dx=\int \frac{u}{1+x^2} dx=\int \frac{u}{\color{red}1+x^2} ({\color{red}1+x^2}) du$ $=\int udu$
Originally Posted by o_O
??? You're stuck? You just substitute it in ...

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

$\int \frac{{\color{red}\arctan x}}{{\color{blue}1 + x^{2}}} {\color{blue} dx} = \int \left({\color{red}\arctan x} \cdot {\color{blue}\frac{1}{1+x^{2}}dx}\right) = \int {\color{red} u} {\color{blue}du}$
Haha...too quick for you all ....just kidding

7. At least we explained why

8. Next time stud ...

Edit: omg moo beat me again.

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