int x^2/(x-2)(x^2+1)dx
Use partial fractions
There exists some A, B, C such that:
x^2/((x-2)(x^2+1)) = A/(x-2) + (Bx+C)/(x^2+1)
Rearrange:
x^2 = A(x^2+1) + (Bx+C)(x-2)
x^2 = (A+B)x^2 + (C-2B)x + (A-2C)
.: A=2C, C=2B, 1=5B
.: B=1/5, C=2/5, A=4/5
Thus:
x^2/((x-2)(x^2+1)) = 4/(5(x-2)) + x/(5(x^2+1)) + 2/(5(x^2+1))
Integrating each fraction term.
(4/5)∫1/(x-2) dx = (4/5)ln(x-2) + C1
(2/5)∫1/(x^2+1) dx = (2/5)arctan(x)+C2
(1/5)∫x/(x^2+1)dx = (1/10)ln(x^2+1)+C3
Summing (and collating constants):
∫x^2/((x-2)(x^2+1)) dx = (4/5)ln(x-2)+(1/10)ln(x^2+1)+(2/5)arctan(x) + C0