Let I = \(\int\frac{x^2}{(x^2+4)(x^2+9)}dx\)
Put x2 = t, we get
∴ \(\frac{x^2}{(x^2+4)(x^2+9)}\)
= \(\frac{t}{(t+4)(t+9)}\)
Now,
\(\frac{t}{(t+4)(t+9)}\) = \(\frac{A}{(t+4)}+\)\(\frac{B}{(t+9)}\)
= \(\frac{A(t+9)+B(t+4)}{(t+4)(t+9)}\)
⇒ t = (A + B)t + (9A + 4B)
Equating the coefficients, we get
A + B = 1,
9A + 4B = 0
Solving above two equations, we get