Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 30

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