Question

Evaluate the following integral:

\[\int\frac{x^2}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)}dx\]

Answer 1

\[\text{Let }I = \int\frac{x^2}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)}dx\]

We express

\[\frac{x^2}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)} = \frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{x^2 + 9}\]

\[ \Rightarrow x^2 = \left( Ax + B \right)\left( x^2 + 9 \right) + \left( Cx + D \right)\left( x^2 + 4 \right)\]

Equating the coefficients of `x^3 , x^2 , x` and constants, we get

\[0 = A + C\text{ and }1 = B + D\text{ and }0 = 9A + 4C\text{ and }0 = 9B + 4D\]

\[\text{or }A = 0\text{ and }B = - \frac{4}{5}\text{ and }C = 0\text{ and }D = \frac{9}{5}\]

\[ \therefore I = \int\left( \frac{- \frac{4}{5}}{x^2 + 4} + \frac{\frac{9}{5}}{x^2 + 9} \right)dx\]

\[ = - \frac{4}{5}\int\frac{1}{x^2 + 4}dx + \frac{9}{5}\int\frac{1}{x^2 + 9} dx\]

\[ = - \frac{4}{5} \times \frac{1}{2} \tan^{- 1} \frac{x}{2} + \frac{9}{5} \times \frac{1}{3} \tan^{- 1} \frac{x}{3} + c\]

\[ = - \frac{2}{5} \tan^{- 1} \frac{x}{2} + \frac{3}{5} \tan^{- 1} \frac{x}{3} + c\]

\[\text{Hence, }\int\frac{x^2}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)}dx = - \frac{2}{5} \tan^{- 1} \frac{x}{2} + \frac{3}{5} \tan^{- 1} \frac{x}{3} + c\]