Calculus of a Single Variable, Chapter 5, 5.2, Section 5.2, Problem 20

int (x^3-4x^2-4x+20)/(x^2-5)dx
To solve, divide the numerator by the denominator (see attached figure).
= int (x - 4 + x/(x^2-5))dx
= int xdx - int4dx + int x/(x^2-5)dx
For the first integral, apply the formula int x^ndx = x^(n+1)/(n+1)+ C.
For the second integral, apply the formula int adx = ax + C .
= x^2/2 - 4x +C + int x/(x^2-5)dx
For the third integral, apply u-substitution method.
Let
u = x^2-5
Differentiate u.
du=2x dx
(du)/2 =xdx
Plug-in them to the third integral.
=x^2/2 - 4x + C + int 1/(x^2-5)*xdx
=x^2/2 - 4x + C + int 1/u *(du)/2
= x^2/2 - 4x + C + 1/2int 1/u du
Then, apply the formula int 1/xdx = ln|x| + C.
=x^2/2-4x + 1/2ln|u| + C
And, substitute back u = x^2-5 .
=x^2/2 - 4x +1/2ln|x^2-5|+C

Therefore, int (x^3-4x^2-4x + 20)/(x^2-5)dx = x^2/2 - 4x + 1/2ln|x^2-5|+C .