Question

Evaluate the following integral:

\[\int\frac{x^2}{x^4 - x^2 - 12}dx\]

 

Answer 1

\[\text{Let }I = \int\frac{x^2}{x^4 - x^2 - 12}dx\]

We express

\[\frac{x^2}{x^4 - x^2 - 12} = \frac{x^2}{x^4 - 4 x^2 + 3 x^2 - 12}\]

\[ = \frac{x^2}{\left( x^2 - 4 \right)\left( x^2 + 3 \right)}\]
\[ = \frac{A}{x^2 - 4} + \frac{B}{x^2 + 3}\]
\[ \Rightarrow x^2 = A\left( x^2 + 3 \right) + B\left( x^2 - 4 \right)\]

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

\[1 = A + B\text{ and }0 = 3A - 4B\]
\[\text{or }A = \frac{4}{7}\text{ and }B = \frac{3}{7}\]
\[ \therefore I = \int\left( \frac{\frac{4}{7}}{x^2 - 4} + \frac{\frac{3}{7}}{x^2 + 3} \right)dx\]
\[ = \frac{4}{7}\int\frac{1}{x^2 - 4}dx + \frac{3}{7}\int\frac{1}{x^2 + 3} dx\]
\[ = \frac{4}{7} \times \frac{1}{4}\log\left| \frac{x - 2}{x + 2} \right| + \frac{\sqrt{3}}{7} \tan^{- 1} \frac{x}{\sqrt{3}} + c\]
\[ = \frac{1}{7}\log\left| \frac{x - 2}{x + 2} \right| + \frac{\sqrt{3}}{7} \tan^{- 1} \frac{x}{\sqrt{3}} + c\]
\[\text{Hence, }\int\frac{x^2}{x^4 - x^2 - 12}dx = \frac{1}{7}\log\left| \frac{x - 2}{x + 2} \right| + \frac{\sqrt{3}}{7} \tan^{- 1} \frac{x}{\sqrt{3}} + c\]