# Evaluation of the integral $\int\frac{dx}{(e^x+1)(2e^x+3)}$.

Evaluate: $$\int\frac{dx}{(e^x+1)(2e^x+3)}$$

Do I have to apply method of difference or something else?

Use the substitution $t=e^x$, $dt=e^x\ dx=t dx$, and you get a rational integral

$$\int \frac{dx}{(e^x+1)(2e^x+3)}=\int\frac{dt}{t(t+1)(2t+3)}$$

Setting $$t=e^x$$ then we have $$dt=tdx$$ or $$dx=\frac{dt}{t}$$ and our integral is $$\int\frac{dt}{t(t+1)(2t+3)}$$ can you integrate it now? the solution should be $$-\ln \left( t+1 \right) +2/3\,\ln \left( 2\,t+3 \right) +1/3\,\ln \left( t \right)$$ with $t=e^x$

• Perhaps add a further hint in case he gets stuck Feb 24 '17 at 14:47

An easy way will be substitute $u = \frac {1}{2e^x+3}$. Then, $e^x = \frac {1/u-3}{2}$. Thus, we get, $$I = -2\int \frac {u}{(u-1)(3u-1)} \mathrm {d}u$$

Now you can use partial fraction decomposition to simplify $I$. Hope you can take it from here.

• Sorry, my bad XD Feb 24 '17 at 14:52
• @SimplyBeautifulArt No problem.
– user371838
Feb 24 '17 at 14:53