Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 32

inte^x/((e^(2x)+1)(e^x-1))dx
Apply integral substitution:u=e^x
=>du=e^xdx
=int1/((u^2+1)(u-1))du
Now let's create partial fraction template for the integrand,
1/((u^2+1)(u-1))=A/(u-1)+(Bu+C)/(u^2+1)
Multiply the equation by the denominator,
1=A(u^2+1)+(Bu+C)(u-1)
=>1=Au^2+A+Bu^2-Bu+Cu-C
=>1=(A+B)u^2+(-B+C)u+A-C
Equating the coefficients of the like terms,
A+B=0 -------------------------(1)
-B+C=0 -----------------------(2)
A-C=1 -----------------------(3)
Now we have to solve the above three linear equations to get A, B and C,
From equation 1, B=-A
Substitute B in equation 2,
-(-A)+C=0
=>A+C=0 ---------------------(4)
Add equations 3 and 4,
2A=1
=>A=1/2
B=-A=-1/2
Plug in the value of A in equation 4,
1/2+C=0
=>C=-1/2
Plug in the values of A,B and C in the partial fraction template,
1/((u^2+1)(u-1))=(1/2)/(u-1)+((-1/2)u+(-1/2))/(u^2+1)
=1/(2(u-1))-(1(u+1))/(2(u^2+1))
=1/2[1/(u-1)-(u+1)/(u^2+1)]
int1/((u^2+1)(u-1))du=int1/2[1/(u-1)-(u+1)/(u^2+1)]du
Take the constant out,
=1/2int(1/(u-1)-(u+1)/(u^2+1))du
Apply the sum rule,
=1/2[int1/(u-1)du-int(u+1)/(u^2+1)du]
=1/2[int1/(u-1)du-int(u/(u^2+1)+1/(u^2+1))du]
Apply the sum rule for the second integral,
=1/2[int1/(u-1)du-intu/(u^2+1)du-int1/(u^2+1)du] ------------------(1)
Now let's evaluate each of the above three integrals separately,
int1/(u-1)du
Apply integral substitution:v=u-1
dv=du
=int1/vdv
Use the common integral:int1/xdx=ln|x|
=ln|v|
Substitute back v=u-1
=ln|u-1| -------------------------------------------(2)
intu/(u^2+1)du
Apply integral substitution:v=u^2+1
dv=2udu
int1/v(dv)/2
Take the constant out and use standard integral:int1/xdx=ln|x|
=1/2ln|v|
Substitute back v=u^2+1
=1/2ln|u^2+1| ----------------------------------------(3)
int1/(u^2+1)du
Use the common integral:int1/(x^2+a^2)dx=1/aarctan(x/a)
=arctan(u) ------------------------------------------(4)
Put the evaluation(2 , 3 and 4) of all the three integrals in (1) ,
=1/2[ln|u-1|-1/2ln|u^2+1|-arctan(u)]
Substitute back u=e^x and add a constant C to the solution,
=1/2[ln|e^x-1|-1/2ln|e^(2x)+1|-arctan(e^x)]+C