Evaluate $\displaystyle \int \dfrac{x^2}{\left(2 x^2 + 1\right) \left(x^2 - 1\right)} \; dx$
Let $I = \displaystyle \int \dfrac{x^2}{\left(2 x^2 + 1\right) \left(x^2 - 1\right)} \; dx$ $\;\;\; \cdots \; (1)$
Let $x^2 = p$ (change of variable; not substitution)
Then, $\dfrac{x^2}{\left(2 x^2 + 1\right) \left(x^2 - 1\right)} = \dfrac{p}{\left(2 p + 1\right) \left(p - 1\right)}$ $\;\;\; \cdots \; (2)$
Let $\dfrac{p}{\left(2 p + 1\right) \left(p - 1\right)} = \dfrac{A}{2 p + 1} + \dfrac{B}{p - 1}$ $\;\;\; \cdots \; (3)$
$\implies$ $p = A \left(p - 1\right) + B \left(2 p + 1\right)$
When $p = \dfrac{-1}{2}$,
$\dfrac{-1}{2} = \dfrac{-3}{2}A$ $\implies$ $A = \dfrac{1}{3}$
When $p = 1$,
$1 = 3 B$ $\implies$ $B = \dfrac{1}{3}$
Substituting the values of A and B in equation $(3)$ gives
$\dfrac{p}{\left(2 p + 1\right) \left(p - 1\right)} = \dfrac{1/3}{2 p + 1} + \dfrac{1/3}{p - 1}$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ In view of equation $(4)$, equation $(2)$ becomes
$\dfrac{x^2}{\left(2 x^2 + 1\right) \left(x^2 - 1\right)} = \dfrac{1/3}{2 x^2 + 1} + \dfrac{1/3}{x^2 - 1}$ $\;\;\; \cdots \; (5)$
$\therefore$ $\;$ We have from equations $(5)$ and $(1)$
$\begin{aligned}
I & = \dfrac{1}{3} \int \dfrac{dx}{2 x^2 + 1} + \dfrac{1}{3} \int \dfrac{dx}{x^2 - 1} \\\\
& = \dfrac{1}{6} \int \dfrac{dx}{x^2 + \left(1/ \sqrt{2}\right)^2} + \dfrac{1}{3} \int \dfrac{dx}{x^2 - \left(1\right)^2} \\\\
& = \dfrac{1}{6} \times \sqrt{2} \tan^{-1} \left(x \sqrt{2}\right) + \dfrac{1}{3} \times \dfrac{1}{2} \log \left|\dfrac{x - 1}{x + 1}\right| + c \\\\
& = \dfrac{1}{3 \sqrt{2}} \tan^{-1} \left(x \sqrt{2}\right) + \dfrac{1}{6} \log \left|\dfrac{x - 1}{x + 1}\right| + c
\end{aligned}$
$\left[\begin{aligned}
\text{Note: } & \\\\
& \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right) + c \\\\
& \int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a} \log \left|\dfrac{x - a}{x + a}\right| + c
\end{aligned}\right]$