Question

Solve $\int \frac{1}{(x^2+a^2)(x^2+b^2)}dx$.

Answer 1

$\int \frac{dx}{(x^2+a^2)(x^2+b^2)}$

$=\frac{1}{(b^2-a^2)}\int \frac{(x^2+b^2-a^2-x^2)dx}{(x^2+a^2)(x^2+b^2)}$

$=\frac{1}{(b^2-a^2)}\int (\frac{1}{x^2+a^2}-\frac{1}{x^2+b^2})dx$

$=\frac{1}{(b^2-a^2)}[\int \frac{dx}{x^2+a^2}-\int \frac{dx}{x^2+b^2}]$

$=\frac{1}{(b^2-a^2)}[\frac{1}{a}ta{n}^{-}1\left(\frac{x}{a}\right)-\frac{1}{b}ta{n}^{-1}\left(\frac{x}{b}\right)]+C$

$\therefore \int \frac{dx}{(x^2+a^2)(x^2+b^2)}=\frac{1}{(b^2-a^2)}[\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)-\frac{1}{b}ta{n}^{-1}\left(\frac{x}{b}\right)]+C$