Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 29

int(sec^2(x))/(tan^2(x)+5tan(x)+6)dx
Let's apply integral substitution:u=tan(x)
=>du=sec^2(x)dx
=int1/(u^2+5u+6)du
Now we have to write down integrand as sum of partial fraction function, but first we will have to factor the denominator,
1/(u^2+5u+6)=1/(u^2+2u+3u+6)
=1/(u(u+2)+3(u+2))
=1/((u+2)(u+3))
Now let's create partial fraction template,
1/((u+2)(u+3))=A/(u+2)+B/(u+3)
Multiply the above equation by the denominator,
=>1=A(u+3)+B(u+2)
1=Au+3A+Bu+2B
1=(A+B)u+3A+2B
Equating the coefficients of the like terms,
A+B=0 -----------------(1)
3A+2B=1 -----------------(2)
From equation 1:A=-B
Substitute A in equation 2,
3(-B)+2B=1
-3B+2B=1
=>B=-1
Plug in the values in the partial fraction template,
1/((u+2)(u+3))=1/(u+2)-1/(u+3)
int1/(u^2+5u+6)du=int(1/(u+2)-1/(u+3))du
Apply the sum rule,
=int1/(u+2)du-int1/(u+3)du
Use the common integral:int1/xdx=ln|x|
=ln|u+2|-ln|u+3|
Substitute back u=tan(x)
and add a constant C to the solution,
=ln|tan(x)+2|-ln|tan(x)+3|+C