integrate (x^2 + x)/[(x^2-4)(x+4)]not sure where to start, maybe integration by parts?

Answer 1

This one looks like it would benefit from a partial fraction expansion:
x(x+1)/((x+2)(x-2)(x+4)) =
A/(x+2) + B/((x-2) + C/(x+4)

A = 2/(-4)(2) = -1/4
B = 6/(4)(6) = 1/4
C = -12/(-2)(-6) = 1, so you have

(-1/4)(dx/(x+2)) + (1/4)(dx/(x-2)) + (5/3)(dx/(x+4))

The integral is
(-1/4)ln(x+2) + (1/4)ln(x-2) + ln(x+4) + lnC =

ln(C(x+4)(((x-2)/(x+2))^(1/4)))

Answer 2

(x^2 + x)/[(x^2 - 4)(x + 4)]
= [(x^2 - 4) +(x + 4)]/[(x^2 - 4)(x + 4)]
= (x^2 - 4)/[(x^2 - 4)(x + 4)] +(x + 4)/[(x^2 - 4)(x + 4)]
= 1/(x + 4) + 1/(x^2 - 4)

Now integrate, you get: ln(x + 4) + (1/4)*ln[(x - 2)/(x + 2)] + c

Answer 3

integration by parts is quite right:
int(x^2+x)/(x^2-4)(x+4))=
(x^2+x ) (1/(x^2-4) int(1/(x+4)) +1/(x+4) int(x^2-4))+
1/((x^2-4)(x+4)) int(x^2+x)

Answer 4

x^5 + 5x^4 - 20x^2 - 16x