Question

Evaluate the following integral:

\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]

Answer 1

\[\text{Let }I = \int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]
We express
\[\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)} = \frac{A}{x^2 + a^2} + \frac{B}{x^2 + b^2}\]
\[ \Rightarrow x^2 = A\left( x^2 + b^2 \right) + B\left( x^2 + a^2 \right)\]
Equating the coefficients of `x^2` and constants, we get
\[1 = A + B\text{ and }0 = b^2 A + a^2 B\]
\[or A = - \frac{a^2}{b^2 - a^2}\text{ and }B = \frac{b^2}{b^2 - a^2}\]
\[ \therefore I = \int\left( \frac{- \frac{a^2}{b^2 - a^2}}{x^2 + a^2} + \frac{\frac{b^2}{b^2 - a^2}}{x^2 + b^2} \right)dx\]
\[ = - \frac{a^2}{b^2 - a^2}\int\frac{1}{x^2 + a^2}dx + \frac{b^2}{b^2 - a^2}\int\frac{1}{x^2 + b^2} dx\]
\[ = - \frac{a}{b^2 - a^2} \tan^{- 1} \frac{x}{a} + \frac{b}{b^2 - a^2} \tan^{- 1} \frac{x}{b} + c\]
\[\text{Hence, }\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx = - \frac{a}{b^2 - a^2} \tan^{- 1} \frac{x}{a} + \frac{b}{b^2 - a^2} \tan^{- 1} \frac{x}{b} + c\]