How do you integrate int e^x/[(e^x-2)(e^(2x)+1)]dx using partial fractions?

1 Answer
Nov 10, 2016

1/5(lnabs(e^x-2)-1/2lnabs(e^(2x)+1)-2 arctan e^x)+c

Explanation:

Substitution with e^x=y then partial fractions

x=lny\ \ , \ \ dx=1/ydy

int1/((y-2)(y^2+1))dy

Search A,B,C so that

1/((y-2)(y^2+1))=A/(y-2)+(By+C)/(y^2+1)
1/((y-2)(y^2+1))=((A+B)y^2+(C-2B)y+A-2C)/((y-2)(y^2+1))

So A=-B, C=2B and A-2C=-B-4B=1

Then B=-1/5,A=1/5, C=-2/5

int1/((y-2)(y^2+1))dy=
=1/5(int1/(y-2)dy-1/2int(2y)/(y^2+1)dy-2int1/(y^2+1)dy)=
=1/5(lnabs(y-2)-1/2lnabs(y^2+1)-2 arctan y)+c

Then substitute y with e^x