Calc II Question

[Moshi2]

hope this helps... Paint lol

the answer will be like what zipykido got

[Internal]

First factor the bottom, it will be ((x+2)+3)((x+2)-3). Then I forget whats its called but there is a trick where you set it up as:
x/((x+2)^2 - 9) = A/((x+2)+3)) + B / ((x+2)-3)

Then obviously multiply to get rid of the denominators and you end up with the equation:

x = A((x+2)+3) + B((x+2)-3)
which is:
x = A(x+5) + B(x-1)
Use different X values to cancel each letter out like plug in X=-5 first to get rid of A
-5 = B(-5-1)
5/6 = B
Do it for A and you should get A = 1/6

Now integrate
1/6/((x+2)+3) + 5/6/((x+2)-3)

Lol that may be really confusing but ya that is what you should do I believe, I took Calculus this year in school but kind of forget it lol.