How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{1}{1 + e^{x}}$ $1/(1+e^x)$ ?

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Answer 1 · 2015-10-17T16:38:52

$int 1/(1+e^(x))\ dx=x-ln(1+e^(x))+C$ .

This is not really a partial fraction problem, but instead it's just a trick. Write:

$1/(1+e^(x))=(1+e^(x)-e^(x))/(1+e^(x))$

$=(1+e^(x))/(1+e^(x))-e^(x)/(1+e^(x))=1-e^(x)/(1+e^(x))$

This means

$int 1/(1+e^(x))\ dx=int (1-e^(x)/(1+e^(x)))\ dx$

$=x-int e^(x)/(1+e^(x))\ dx$ .

For this last integral, let $u=1+e^(x)$ so that $du=e^(x)\ dx$ and we get

$int e^(x)/(1+e^(x))\ dx=int\ (du)/u=ln|u|+C=ln(1+e^(x))+C$ (note that $1+e^(x)>0$ for all $x$ ).

Therefore,

$int 1/(1+e^(x))\ dx=x-ln(1+e^(x))+C$ .

How do you use partial fraction decomposition to decompose the fraction to integrate 1/(1+e^x) 11+ex1/(1+e^x) ?