Let ${\mathit{x}}^2$ = y
$\frac{{\mathit{x}}^2}{\left({\mathit{x}}^2+4\right)\left({\mathit{x}}^2+9\right)}$ = $\frac{\mathit{y}}{\left(\mathit{y}+\right)\left(\mathit{y}+9\right)}$ = $\frac{\mathit{A}}{\mathit{y}+4}$ + $\frac{\mathit{B}}{\mathit{y}+9}$
y = A (y + 9) + B (y + 4)
Comparing both sides,
A + B = 1 and 9A + 4B = 0
Solving, we get A = $\frac{-4}{5}$ and B = $\frac{9}{5}$
∴ I = ∫ $\left|\frac{-4}{5\left({\mathit{x}}^2+4\right)}\right|+\frac{9}{5\left({\mathit{x}}^2+9\right)}$ dx
= $-\frac{4}{5}\times \frac{1}{2}\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{\mathit{x}}{2}\right)$ + $\frac{9}{5}\times \frac{1}{3}\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{\mathit{x}}{3}\right)$ + X
= - $\frac{2}{5}\mathit{t}\mathit{a}{\mathit{n}}^{-1}\frac{\mathit{x}}{2}+\frac{3}{5}\mathit{t}\mathit{a}{\mathit{n}}^{-1}\frac{\mathit{x}}{3}$ + C