Evaluate the indefinite integral $\int\frac{dx}{(1+e^x)^2}$

Question

Evaluate the indefinite integral $$\int\frac{dx}{(1+e^x)^2}$$

There is some clever trick to solve this, I think.

I'm really hesitant to ask a homework question without submitting an attempted solution, but this question is not very conducive towards partial solutions.

Answer 1

You are given $$I = \int {\frac{{dx}}{{{{\left( {{e^x} + 1} \right)}^2}}}} $$

Let $e^x+1=u$. Then, what does you integral become?

SPOILER You should get

$I = \displaystyle \int {\frac{{du}}{{{u^2}\left( {u - 1} \right)}}}$

Then use partial fraction decomposition.

Answer 2

If you make the change of variables $u=e^x$ then you get

$$\int\frac{1}{(1+e^x)^2}=\int\frac{1}{u(1+u)^2}=-\int\frac{1}{u+1}-\int\frac{1}{(u+1)^2}+\int\frac{1}{u}=$$ $$=\frac{(1+u)(\log{u}-\log{(u+1))+1}}{u+1}$$

Now undo the change of variables, operate an you will get

$\int\frac{1}{(1+e^x)^2}=x+\frac{1}{e^x+1}+\log{(e^x+1)}+K$

Answer 3

$\displaystyle \int\frac{1}{(1+e^x)^2}dx = \int\frac{e^{-x}.e^{-x}}{(e^{-x}+1)^2}dx$

Let $e^{-x}+1=t\Leftrightarrow -e^{-x}dx = dt\Leftrightarrow e^{-x}dx = -dt$

So $\displaystyle -\int \frac{(t-1)}{t^2}dt = \int \frac{1-t}{t^2}dt=\int t^{-2}dt-\int \frac{1}{t}dt$

$\displaystyle = -\frac{1}{t}-\ln \mid t \mid +C$

$\displaystyle = -\frac{1}{e^{-x}+1}-\ln \mid e^{-x}+1 \mid+C$