Hi, can you think of a way to solve the following integral using substitution?
$\displaystyle \int{e^{-x}\arctan{e^x}}dx$
Help much appreciated.
oh, did i make a mistake?
we have $\displaystyle \int \frac {\arctan e^x}{e^x}~dx$
$\displaystyle = \int \frac {e^x \arctan e^x}{e^{2x}}~dx$
Let $\displaystyle u = e^x$
$\displaystyle \Rightarrow du = e^x~dx$
So our integral becomes
$\displaystyle \int \frac {\arctan u}{u^2}~du$
Start by integrating by parts:
$\displaystyle \int {e^{ - x} \arctan e^x \,dx} = - e^{ - x} \arctan e^x + \underbrace {\int {\frac{1}
{{e^{2x} + 1}}\,dx} }_\lambda .$
Now $\displaystyle \lambda$ can be evaluated as follows:
$\displaystyle \lambda = \int {\frac{{e^{2x} + 1 - e^{2x} }}
{{e^{2x} + 1}}\,dx} = x - \frac{1}
{2}\ln \left( {e^{2x} + 1} \right) + k .$
The full answer is
$\displaystyle \int {e^{ - x} \arctan e^x \,dx} = - e^{ - x} \arctan e^x + x - \frac{1}
{2}\ln \left( {e^{2x} + 1} \right) + k.$