How do you use partial fractions to find the integral $int (e^x)/((e^x-1)(e^x+4))dx$ ?

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Answer 1 · 2017-07-20T07:33:24

$inte^x/((e^x-1)(e^x+4))dx=1/5(ln|1-e^(-x)|-ln|1+4e^(-x)|)$

Let $e^x/((e^x-1)(e^x+4))=A/(e^x-1)+B/(e^x+4)$

$=(Ae^x+4A+Be^x-B)/((e^x-1)(e^x+4))=(e^x(A+B)+4A-B)/((e^x-1)(e^x+4))$

Comparing like terms, $4A-B=0$ i.e. $B=4A$ and $A+B=1$

i.e. $A=1/5$ and $B=4/5$ and

$inte^x/((e^x-1)(e^x+4))dx=1/5int1/(e^x-1)dx+4/5int1/(e^x+4)dx$

= $1/5inte^(-x)/(1-e^(-x))dx+4/5inte^(-x)/(1+4e^(-x))dx$

Let $u=1-e^(-x)$ then $du=e^(-x)dx$ and

$v=1+4e^(-x)$ then $dv=-4e^(-x)dx$

Hence $1/5inte^(-x)/(1-e^(-x))dx+4/5inte^(-x)/(1+4e^(-x))dx$

= $1/5int(du)/u-1/5int(dv)/v$

= $1/5(ln|u|-ln|v|)$

= $1/5(ln|1-e^(-x)|-ln|1+4e^(-x)|)$

How do you use partial fractions to find the integral int (e^x)/((e^x-1)(e^x+4))dxint (e^x)/((e^x-1)(e^x+4))dx?