Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 31

inte^x/((e^x-1)(e^x+4))dx
Let's apply integral substitution:u=e^x
=>du=e^xdx
=int1/((u-1)(u+4))du
Now create partial fraction template of the integrand,
1/((u-1)(u+4))=A/(u-1)+B/(u+4)
Multiply the above equation by the denominator,
=>1=A(u+4)+B(u-1)
1=Au+4A+Bu-B
1=(A+B)u+4A-B
Equating the coefficients of the like terms,
A+B=0 ----------------(1)
4A-B=1 ----------------(2)
From equation 1, A=-B
Substitute A in equation 2,
4(-B)-B=1
-5B=1
B=-1/5
A=-B=-(-1/5)
A=1/5
Plug in the values of A and B in the partial fraction template,
1/((u-1)(u+4))=(1/5)/(u-1)+(-1/5)/(u+4)
=1/(5(u-1))-1/(5(u+4))
int1/((u-1)(u+4))du=int(1/(5(u-1))-1/(5(u+4)))du
Apply the sum rule,
=int1/(5(u-1))du-int1/(5(u+4))du
Take the constant out,
=1/5int1/(u-1)du-1/5int1/(u+4)du
Now use the common integral:int1/xdx=ln|x|
=1/5ln|u-1|-1/5ln|u+4|
Substitute back u=e^x and add a constant C to the solution,
=1/5ln|e^x-1|-1/5ln|e^x+4|+C