After trying this multiple ways, I give up. Here's the integral:
$$\int e^{-x}\arctan(e^x)\,dx$$
I have set
$u=\arctan(e^x)$ and $dv=e^{-x}d\,x$
and have obtained $du=\dfrac{e^x \, dx}{1+e^{2x}}$ and $v=-e^{-x}$
Using the integration by parts formula,
$$\int u\, dv = uv - \int v\,du$$
I got $-e^{-x}\arctan(e^x)+\int \dfrac{1}{1+e^{2x}}\,dx$
How would I solve $\int \dfrac{1}{1+e^{2x}}\,dx$ ? That's the part I'm stuck on.
On
In $I = \int e^{-x}\arctan(e^x)\,dx$, let $y = e^{-x}$.
Since $dy = -e^{-x} dx$, $I = \int -dy \arctan(1/y) =-\int \arctan(\pi/2-y)\, dy =\int \arctan(y-\pi/2)\, dy $
and then use $\int \arctan x\, dx = x \arctan x - (1/2) ln(1+x^2) + C $.
On
Everyone's responses were very helpful, and after some further tinkering, I'd like to add my own.
We can pull out the $e^{2x}$ in the denominator of
$$\int \dfrac{1}{1+e^{2x}}\,dx$$
and obtain $$\int \dfrac{1}{e^{2x}(1+e^{-2x})}\,dx$$
which simplifies to $$\int \dfrac{e^{-2x}}{1+e^{-2x}}\,dx.$$
We can now use u-substitution by setting $$u=1+e^{-2x}$$
and, by taking the derivative of $u$, we get $$du=-2e^{-2x}dx.$$
The integral now becomes $$-\dfrac{1}{2}\int\dfrac{du}{u}$$
which yields $$-\frac12\ln{(1+e^{-2x})}+C.$$
Once again, thank you all for your kind help. It was greatly appreciated.
$$\int \frac{dx}{1+e^{2 x}} = \int dx \frac{e^{-x}}{e^x+e^{-x}} = - \int du \frac{1}{u+u^{-1}}$$
which is
$$-\int du \frac{u}{1+u^2} = -\frac12 \log{(1+u^2)} + C = -\frac12 \log{\left (1+e^{-2 x}\right)} + C$$