Question

Evaluate: $\int \frac{{\text{x}}^2}{{\text{x}}^4-{\text{x}}^2-12}\text{dx}$

Answer 1

$\int \frac{{\text{x}}^2}{{\text{x}}^4-{\text{x}}^2-12}\text{dx}$
$=\int \frac{{\text{x}}^2}{{\text{x}}^4-{\text{4x}}^2+{\text{3x}}^2-12}\text{dx}$
$=\int \frac{{\text{x}}^2}{{\text{(x}}^2-\text{4)}{\text{(x}}^2+3)}\text{dx}$
Substituting ${\text{x}}^2=\text{t}$ for partial fraction
$\therefore \frac{{\text{x}}^2}{({\text{x}}^2-4)({\text{x}}^2+3)}=\frac{\text{t}}{(\text{t}-4)(\text{t}+3)}$
Let $\frac{\text{t}}{(\text{t}-4)(\text{t}+3)}=\frac{\text{A}}{(\text{t}-4)}+\frac{\text{B}}{(\text{t}+3)}$
$\Rightarrow \text{t}=\text{A}(\text{t}+3)+\text{B}(\text{t}-4)$
Substituting $\text{t}=-3$,
$-3=\text{B}(-3-4)\Rightarrow \text{B}=\frac{3}{7}$
Substituting $\text{t}=4$,
$4=\text{A}(4+3)\Rightarrow \text{A}=4/7$
So, $\frac{\text{t}}{(\text{t}-4)(\text{t}+3)}=\frac{1}{7}[\frac{4}{(\text{t}-4)}+\frac{3}{(\text{t}+3)}]$
$\therefore \int \frac{{\text{x}}^2}{({\text{x}}^2-4)({\text{x}}^2+3))}\text{dx}$
$=\frac{1}{7}\int [\frac{4}{({\text{x}}^2-4)}+\frac{3}{({\text{x}}^2+3)}]\text{dx}$
$=\frac{1}{7}[4\int \frac{1}{{\text{x}}^2-2^2}\text{dx}+3\int \frac{1}{{\text{x}}^2+(\surd 3{)}^2}\text{dx}]$
$=\frac{1}{7}[\frac{4}{4}\text{log}\left|\frac{\text{x}-2}{\text{x}+2}\right|+\frac{3}{\sqrt{3}}{\text{tan}}^{-1}\left(\frac{\text{x}}{\surd 3}\right)]+\text{c}$
$[\because \int \frac{\text{dx}}{({\text{x}}^2-{\text{a}}^2)}=\frac{1}{\text{2a}}log\left|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\right|+\text{c},\int \frac{1}{({\text{x}}^2+{\text{a}}^2)}\text{dx}=\frac{1}{\text{a}}{\text{tan}}^{-1}(\frac{\text{x}}{\text{a}})+\text{c}]$
$=\frac{1}{7}\text{log}\left|\frac{\text{x}-2}{\text{x}+2}\right|+\frac{\sqrt{3}}{7}{\text{tan}}^{-1}\left(\frac{\text{x}}{\sqrt{3}}\right)+\text{c}$
Where c is constant of integration
Hence, the integration is
$\frac{1}{7}\text{log}\left|\frac{\text{x}-2}{\text{x}+2}\right|+\frac{\sqrt{3}}{7}{\text{tan}}^{-1}\left(\frac{\text{x}}{\sqrt{3}}\right)+\text{c}$