Question

Solve the integral: 5/x(x^2-4) dx

I've done the partial fractions and found B and C, both equaling 5/8; and that's gonna give me 5/8 ln|x^2-4|. I just need help finding A

Answer 1

A = -5/4

where did you get stuck?

Answer 2

I think I got it. You just make x=0 to find A. Thanks.

Answer 3

let 5/x(x^2-4) = A/x + B/(x+2) + C/(x-2)
= [ A(x^2 - 4) + Bx(x-2) + Cx(x+2)]/(x(x^2 - 4))

then A(x^2 - 4) + Bx(x-2) + Cx(x+2) = 5 for all values of x
let x = 0 ----> -4A + 0 + 0 = 5 , so A = -5/4
let x = 2 ----> 0 + 0 + 8C = 5 , so C = 5/8
let x = -2 ---> 0 -8B + 0 = 5, so B = -5/8

∫ 5/x(x^2-4) dx = ∫ ( (-5/4)/x - (5/8)/(x+2) + (5/8) /(x-2) )dx
= (-5/4) lnx - (5/8) ln(x+2) + (5/8) ln (x-2) + c
= (-5/8)(2 lnx - ln(x+2) - ln(x-2) ) + c