$\int_0^\infty \frac{x^2}{(x^2-4)(x^2-9)}\,\text dx$

Question

I am trying to compute the following contour integration but am quite stuck I have to evaluate it analytically, by extending it to the complex plane and solving an appropriate integral involving a complex variable $z= x + iy$:

$$\int_0^\infty \frac{x^2}{(x^2-4)(x^2-9)}\,\text dx$$

Answer 1

HINT:

W/O using Complex Calculus,

$$\frac{x^2}{(x^2-4)(x^2-9)}=\frac15\cdot\frac{9(x^2-4)-4(x^2-9)}{(x^2-4)(x^2-9)}=\cdots$$

Now use this

Answer 2

by using the partial fraction $$\int_{0}^{\infty }\frac{x^2}{(x^2-4)(x^2-9)}dx=(\int_{0}^{\infty }\frac{1}{5(x-2)}-\frac{1}{5(x-2)}+\frac{3}{10(x-3)}-\frac{3}{10(x+3)})dx$$ $$=\frac{1}{5}\log(\frac{x-2}{x+2})+\frac{3}{10}\log(\frac{x+3}{x-3}) $$ take the limit when $x=\infty$ to get $$\lim_{x\rightarrow \infty }\log(\frac{x-2}{x+2})=0$$ $$\lim_{x\rightarrow \infty }\log(\frac{x+3}{x-3})=0$$

$$I=-\frac{1}{5}\log(-1)-\frac{3}{10}\log(-1)=-\frac{i\pi }{2}$$