Question

Evaluate the following integral:

\[\int\frac{x^2}{1 - x^4}dx\]

Answer 1

\[Let I = \int\frac{x^2}{1 - x^4}dx\]

We express

\[\frac{x^2}{1 - x^4} = \frac{x^2}{\left( 1 - x^2 \right)\left( 1 + x^2 \right)}\]
\[ = \frac{A}{1 - x^2} + \frac{B}{1 + x^2}\]
\[ \Rightarrow x^2 = A\left( 1 + x^2 \right) + B\left( 1 - x^2 \right)\]

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

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