Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 11

int(2x^3-4x^2-15x+5)/(x^2-2x-8)dx
The integrand is a improper rational function,as the degree of the numerator is greater than the degree of the denominator.So we have to carry out division.
(2x^3-4x^2-15x+5)/(x^2-2x-8)=2x+(x+5)/(x^2-2x-8)
Since the polynomials do not completely divide, we have to continue with the partial fractions on the remainder and factor out the denominator.
(x+5)/(x^2-2x-8)=(x+5)/(x^2-4x+2x-8)
=(x+5)/(x(x-4)+2(x-4))
=(x+5)/((x-4)(x+2))
Now we create the partial fraction template,
(x+5)/((x-4)(x+2))=A/(x-4)+B/(x+2)
Multiply the above equation with the denominator,
=>(x+5)=A(x+2)+B(x-4)
=>x+5=Ax+2A+Bx-4B
=>x+5=(A+B)x+2A-4B
Equating the coefficients of the like terms,
A+B=1 ---------------(1)
2A-4B=5 -------------(2)
Now we to solve the above linear equations to get the values of A and B,
Multiply equation 1 by 4,
4A+4B=4 -------------(3)
Now add equation 2 and 3,
2A+4A=5+4
=>6A=9
=>A=9/6
=>A=3/2
Plug in the value of A in equation 1 ,
3/2+B=1
=>B=1-3/2
=>B=-1/2
Plug in the values of A and B in the partial fraction template,
(x+5)/((x-4)(x+2))=(3/2)/(x-4)+(-1/2)/(x+2)
=3/(2(x-4))-1/(2(x+2))
int(2x^3-4x^2-15x+5)/(x^2-2x-8)dx=int(2x+3/(2(x-4))-1/(2(x+2)))dx
Apply the sum rule,
=int2xdx+int3/(2(x-4))dx-int1/(2(x+2))dx
Take the constant out,
=2intxdx+3/2int1/(x-4)dx-1/2int1/(x+2)dx
Apply the power rule for the first integral and use the common integral int1/xdx=ln|x| for the second and third integral:
=2x^2/2+3/2ln|x-4|-1/2ln|x+2|
Simplify and a constant C to the solution,
=x^2+3/2ln|x-4|-1/2ln|x+2|+C