Question

use the method of equating coefficients.

integral of (x^3+5x^2+12)/((x^2)(x^2+4)) dx

Answer 1

(x^3+5x^2+12)/((x^2)(x^2+4)) = (x+2)/(x^2+4) + 3/x^2

So,

∫(x^3+5x^2+12)/((x^2)(x^2+4))
= ∫ (x+2)/(x^2+4) + 3/x^2 dx
= 1/2 log(x^2+4) + arctan(x/2) - 3/x + C

Answer 2

equating coefficients = partial fractions ?

let (x^3+5x^2+12)/((x^2)(x^2+4))
= A/x^2 + (Bx+C)/(x^2+)
= (Ax^2 + 4A + Bx^3 + Cx^2)/(x^2(x^2+4))
= (Bx^3 + (A+C)x^2 + 4A)/(x^2(x^2+4))

so B=1
A+C=5
4A=12 ---> A = 3
in A+C=5 ---> C=2

so we can write:
(x^3+5x^2+12)/((x^2)(x^2+4))
=3/x^2 +(x+2)/(x^2+4)

and ∫(x^3+5x^2+12)/((x^2)(x^2+4)) dx
= ∫ 3/x^2 dx + ∫ (x+2)/(x^2+4)dx
= -3/x + (1/2)ln(x^2+4) + PPPPP

where PPPP = ∫ 4/(x^2+4) dx
from my formula tables from "years ago" this is
tan^-1 (x/2)

= -3/x + (1/2)ln(x^2+4) + tan^-1 (x/2) + constant

better check this, I did not write it out on paper first.