1. ## integration by substitution

Hi, can you think of a way to solve the following integral using substitution?

$\int{e^{-x}\arctan{e^x}}dx$

Help much appreciated.

2. Originally Posted by disclaimer
Hi, can you think of a way to solve the following integral using substitution?

$\int{e^{-x}\arctan{e^x}}dx$

Help much appreciated.
i don't think substitution alone will work.

we can begin with substitution.

a substitution of $u = e^x$ yields $\int \frac {\arctan u}{u^2}~du$

which we can do by parts

3. Originally Posted by Jhevon
i don't think substitution alone will work.

we can begin with substitution.

a substitution of $u = e^x$ yields $\int \frac {\arctan u}{u^2}~du$

which we can do by parts
How did you use the substitution to get $\int \frac {\arctan u}{u^2}~du$ ?

4. Originally Posted by wingless
How did you use the substitution to get $\int \frac {\arctan u}{u^2}~du$ ?
oh, did i make a mistake?

we have $\int \frac {\arctan e^x}{e^x}~dx$

$= \int \frac {e^x \arctan e^x}{e^{2x}}~dx$

Let $u = e^x$

$\Rightarrow du = e^x~dx$

So our integral becomes

$\int \frac {\arctan u}{u^2}~du$

5. Ah, OK, my bad.. I saw the integral wrong..

6. Originally Posted by disclaimer
$\int{e^{-x}\arctan{e^x}}dx$
Start by integrating by parts:

$\int {e^{ - x} \arctan e^x \,dx} = - e^{ - x} \arctan e^x + \underbrace {\int {\frac{1}
{{e^{2x} + 1}}\,dx} }_\lambda .$

Now $\lambda$ can be evaluated as follows:

$\lambda = \int {\frac{{e^{2x} + 1 - e^{2x} }}
{{e^{2x} + 1}}\,dx} = x - \frac{1}
{2}\ln \left( {e^{2x} + 1} \right) + k .$

$\int {e^{ - x} \arctan e^x \,dx} = - e^{ - x} \arctan e^x + x - \frac{1}