Evaluate the integral of : $\int \:\frac{e^{2x}}{1+e^{3x}}dx$

Evaluate the integral of : $\int \:\frac{e^{2x}}{1+e^{3x}}dx$

So I start by making : $e^{2x}=t$

and this gives me the following : $\frac{1}{2}\int \:\frac{dt}{1+t\sqrt{t}}$

Then using symbolab calculator I got :

https://www.symbolab.com/solver/definite-integral-calculator/%5Cint_%7B%20%7D%5E%7B%20%7D%20%5Cfrac%7B1%7D%7B1%2Bx%5Csqrt%7Bx%7D%7Ddx/?origin=enterkey

I would like to know if there is a fastest way to solve it.

• Why not try $t=e^x$? – carmichael561 May 1 '16 at 20:18
• @carmichael561 thanks in this way is easier – GiovanS May 1 '16 at 20:23

Integrals of the form $$\int\frac{p(e^x)}{q(e^x)}\,dx$$ where $p$ and $q$ are polynomials can always be reduced to rational functions with the substitution $t=e^x$, that brings the integral to the form $$\int\frac{p(t)}{tq(t)}\,dt$$ owing to the fact that $x=\log t$, so $dx=\frac{1}{t}\,dt$.
In your case you get $$\int\frac{t}{1+t^3}\,dt$$ that can be computed by partial fractions: $$\frac{t}{1+t^3}=\frac{a}{1+t}+\frac{bt+c}{1-t+t^2}$$ and standard techniques afterwards.