Integrate the function: $\frac{1}{(x^{2}+1)(x^{2}+4)}$

Let $\frac{1}{(x^{2}+1)(x^{2}+4)} = \frac{A}{x^{2}+1} + \frac{B}{x^{2}+4}$.
Substituting $y = x^{2}$,we have $\frac{1}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4}$.
$1 = A(y+4) + B(y+1)$.
For $y = -1$,$1 = A(3) \Rightarrow A = \frac{1}{3}$.
For $y = -4$,$1 = B(-3) \Rightarrow B = -\frac{1}{3}$.
Thus,$\frac{1}{(x^{2}+1)(x^{2}+4)} = \frac{1}{3(x^{2}+1)} - \frac{1}{3(x^{2}+4)}$.
Integrating both sides:
$\int \frac{1}{(x^{2}+1)(x^{2}+4)} dx = \frac{1}{3} \int \frac{1}{x^{2}+1} dx - \frac{1}{3} \int \frac{1}{x^{2}+2^{2}} dx$.
Using the standard integrals $\int \frac{1}{x^{2}+a^{2}} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$= \frac{1}{3} \tan^{-1}(x) - \frac{1}{3} \cdot \frac{1}{2} \tan^{-1}(\frac{x}{2}) + C$.
$= \frac{1}{3} \tan^{-1}(x) - \frac{1}{6} \tan^{-1}(\frac{x}{2}) + C$.