what is the integral of (4x^2-2)/(x^2-6x-40)

Answer 1

First do a long division
(4x^2-2)/(x^2-6x-40) = 4 + (24x+158)/(x^2 - 6x - 40)

now separate (24x+158)/(x^2 - 6x - 40
into partial fractions so that
A/(x-10) + B(x+4) = (24x + 158)/((x-10)(x+4))

then
A(x+4) + B(x-10) = 24x + 158
let x = -4
-14B = 62
B = -31/7

let x = 10
14A = 398
A = 199/7

(4x^2-2)/(x^2-6x-40) = 4 + (199/7)/(x-10) - (31/7)/(x+4)

so the integral is
4x + (199/7)ln(x-10) - (31/7)ln(x+10) + C , where C is a constant

Answer 2

I have a typo, the last line should say:

4x + (199/7)ln(x-10) - (31/7)ln(x+4) + C , where C is a constant