Evaluating $\int{\frac{du}{3e^{u}+1}}$

Question

why is $$\int{\frac{du}{3e^{u}+1}}=\ln\frac{e^u}{3e^u+1}+c$$ ? I think some substitution should help solving this integral, but everything I tried did not work.

Answer 1

Hint:

$$\ \frac{1}{3e^u+1}=\frac{1}{e^u(3+e^{-u})}=\frac{e^{-u}}{3+e^{-u}}$$

and now substitution

$$\ e^{-u}=t, -e^{-u}du=dt$$

Answer 2

Let $x=e^u$, then you will have $du=\frac{dx}{x}$, thus your integral will reduces to $$\int{\frac{dx}{x(3x+1)}}=\int{\frac{dx}{x}}-\int{\frac{3}{3x+1}dx}=\ln{x}-\ln{(3x+1)}+C=\ln{\frac{x}{3x+1}}+C$$ Thus the solution is $\ln{\frac{e^u}{3e^u+1}}+C$

Answer 3

You can use $\displaystyle\int\frac{du}{3e^{u}+1}=\int\frac{(3e^{u}+1)-3e^{u}}{3e^{u}+1}du=\int(1-\frac{3e^{u}}{3e^{u}+1})du$

$\;\;\;\;\;\displaystyle=[u-\ln(3e^{u}+1)]+C=\ln\left(\frac{e^u}{3e^{u}+1}\right)+C$