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

1 Answer
Jul 20, 2017

#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)|)#