Question

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

Answer 1

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

Explanation:

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)|)`