Hi, I have a final in a few hours and I can't figure out how to integrate this by partial fractions:

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


Any help would be much appreciated, thanks.

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

You know its gonna be of the form :

(2x + 1) / (x-4)(x^2 + 1) = A/(x-4) + (Bx + C)/(x^2 +1)
(2x + 1) / (x-4)(x^2 + 1) = (A(x^2 +1) + (Bx + C)(x-4) ) / (x-4)(x^2 + 1)
(2x + 1) = A(x^2 +1) + (Bx + C)(x-4)
(2x + 1) = Ax^2 + A + Bx^2 -4Bx + Cx - 4C

by association you see that

1) Ax^2 + Bx^2 = 0 --> A + B = 0
2) -4Bx + Cx = 2x --> -4B +C = 2
3) A - 4C = 1


solve for A, B, C and now you have your seperate fractions of the form A/(x-4) + (Bx + C)/(x^2 +1)

integrate each fraction seperately and add them up

also see : http://www.youtube.com/watch?v=6qVgHWxdlZ0

This post was edited by LeB on Aug 21 2014 03:26pm

Thanks a ton man, that just helped me a lot