How to find $\int {dx\over{e^x(e^x+1)^2}}$

Question

How to find

$$\int {dx\over{e^x(e^x+1)^2}}$$ First I put $x=log y$ and then by putting ${1\over y}=z$ I came to an result. But is there any better alternative?

Answer 1

HINT:

Putting $$e^x=u,e^xdx=du$$

$$\int {dx\over{e^x(e^x+1)^2}}=\int\frac{e^xdx}{(e^x)^2(e^x+1)^2}=\int \frac{du}{u^2(u+1)^2}$$

$$\text{Now, }\frac1{u^2(u+1)^2}=\frac{(u+1-u)^2}{u^2(u+1)^2}=\frac{(u+1)^2+u^2-2u(u+1)}{u^2(u+1)^2}$$ $$=\frac1{u^2}+\frac1{(u+1)^2}-2\frac1{u(u+1)}$$

$$\text{Again, }\frac1{u(u+1)}=\frac{u+1-u}{u(u+1}=\frac1u-\frac1{u+1}$$

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Alternatively, using Partial Fraction decomposition, $$\frac1{u^2(u+1)^2}=\frac Au+\frac B{u^2}+\frac C{1+u}+\frac D{(1+u)^2}$$

where $A,B,C,D$ are arbitrary constants.

Multiply either sides by $u^2(u+1)^2$ and compare the coefficients of the different powers of $u$ to determine $A,B,C,D$