Integral of (2x^3+x^2-8x-1)/(x^2-x-2)

Not sure how to do this problem, can someone please give a step by step answer?

Since the degree of the numerator exceeds that of the denominator, perform division first to get:
(2x^3 + x^2 - 8x - 1)/(x^2 - x - 2) = (2x + 3) - (x - 5)/(x^2 - x - 2).

Then, since the denominator factors to (x - 2)(x + 1), we see that, by Partial Fractions:
(x - 5)/(x^2 - x - 2) = A/(x - 2) + B/(x + 1).

Multiplying both sides by x^2 - x - 2 gives:
x - 5 = A(x + 1) + B(x - 2).

(i) Letting x = 2 ==> 3A = -3 ==> A = -1
(ii) Letting x = -1 ==> -3B = -6 ==> B = 2.

Thus:
(2x^3 + x^2 - 8x - 1)/(x^2 - x - 2)
= (2x + 3) - (x - 5)/(x^2 - x - 2)
= (2x + 3) - [-1/(x - 2) + 2/(x + 1)]
= (2x + 3) + 1/(x - 2) - 2/(x + 1).

Integrating term-by-term yields:
∫ (2x^3 + x^2 - 8x - 1)/(x^2 - x - 2) dx
= ∫ [(2x + 3) + 1/(x - 2) - 2/(x + 1)] dx
= x^2 + 3x + ln|x - 2| - 2ln|x + 1| + C.

I hope this helps!