Evaluate: ∫x²/(x²+4)(x²+9)dx

I=∫x²/(x²+4)(x²+9)dx

Let's use partial fraction decomposition:
x²/(x²+4)(x²+9) = (Ax+B)/(x²+4) + (Cx+D)/(x²+9)
x² = (Ax+B)(x²+9) + (Cx+D)(x²+4)
x² = Ax³ + 9Ax + Bx² + 9B + Cx³ + 4Cx + Dx² + 4D
x² = (A+C)x³ + (B+D)x² + (9A+4C)x + (9B+4D)

Comparing coefficients:
A + C = 0
B + D = 1
9A + 4C = 0
9B + 4D = 0

Solving this system of equations, we get:
A = 0
B = -5/5 = -1
C = 0
D = 6/5

Therefore,
x²/(x²+4)(x²+9) = (-1)/(x²+4) + (6/5)/(x²+9)

I = ∫[(-1)/(x²+4) + (6/5)/(x²+9)]dx
I = -∫1/(x²+4)dx + (6/5)∫1/(x²+9)dx

We know that ∫1/(x²+a²)dx = (1/a)arctan(x/a)

I = -(1/2)arctan(x/2) + (6/5)(1/3)arctan(x/3) + C
I = -(1/2)arctan(x/2) + (2/5)arctan(x/3) + C