Evaluate integral(5x-5/3x^2-8x-3) dx.

Answer 1

Use partial fractions

(5x-5)/(3x^2 - 8x - 3)
= (5x-5)/((x-3)(3x+1))

let (5x-5)/((x-3)(3x+1))
= A/(x-3) + B/(x-3)
= (A(x-3) + B(3x+1))/((x-3)(3x+1))

so

5x - 5 = A(x-3) + B(3x+1)
let x=3 ----> 10 = B(10) ---> B=1
let x = -1/3
-20/3 = A(-10/3) ----> A = 2

so

(5x-5)/((3x^2 - 8x - 3)
= 2/(x-3) + 1/(3x+1)

and the integral of that is
2ln(x-3) + (1/3)ln(3x+1) + c