How can we integrate $\int 1/(e^x + e^{2x}) dx$? This is my first post off my phone, so please try to forgive me for the poor format. I’m attempting to take the antiderivative:
$$ \int \frac{1}{e^x + e^{2x}} dx.$$
And I have no idea how to do so. I’ve only taken up to calc 3 and I can’t think of anything I’ve learned that can solve this, but there is an answer. I’m not so much interested in the answer as I am what method is used to solve this.
 A: Just for fun, here's an approach that avoids doing partial fractions on $1/(u^2+u^3)$ (and then having to resubstitute $u=e^x$):
$${1\over e^x+e^{2x}}={1\over e^x}-{1\over1+e^x}=e^{-x}-{(1+e^x)-e^x\over1+e^x}=e^{-x}-1+{e^x\over1+e^x}$$
Thus
$$\int{1\over e^x+e^{2x}}dx=-e^{-x}-x+\ln(1+e^x)+C$$
(The substitution $u=e^x$ lurks within the part that gives $\ln(1+e^x)$, but it's so easy it doesn't need to be made explicit.)
A: You are trying to integrate
$$ \int \frac{1}{e^x + e^{2x}}dx.$$
There are many ways to do this, but one which jumps out at me quickly is to perform $u$-substitution and set $u = e^x$, so that
$$ \int \frac{1}{e^x + e^{2x}} dx = \int \frac{1}{e^{2x} + e^{3x}} (e^x dx) = \int \frac{1}{u^2(u + 1)} du,$$
and this integral quickly yields to partial fraction decomposition.
A: Yeah so just multiply and divide the whole quantity by $e^{x}$ and put $t=e^{x}$, then the integral becomes $$\int \frac{1}{t^2+t^3} \ dt$$
A: Take $u=e^x$ then we get $e^{-x}du=dx$ and thus 
$$
\int\frac{1}{e^x+e^{2x}}dx=\int\frac{1}{u^2+u^3}du
$$
which you can tackle by partial fractions.
A: \begin{align}
\int \frac{1}{e^x + e^{2x}} dx&=\int \frac{e^{-x}}{1 + e^{x}} dx\\
&=-\int \frac{du}{1 + \frac1u} \tag{By $u=e^{-x}$}\\
&=-\int1-\frac{1}{1+u}du\\
&=-u+\ln(1+u)+C\\
&=-e^{-x}+\ln(1+e^{-x})+C
\end{align}
