x^2 x^4-x^2-12 dx - Vidyadip

Question

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

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Answer

Let $\text{I}=\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\text{dx}$
$=\int\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}\text{dx}$
Now, $\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}=\frac{\text{A}}{\text{x}^2-4}+\frac{\text{B}}{\text{x}^2+3}$
Let $\text{x}^2=\text{t}$
$\Rightarrow\ \frac{\text{t}}{(\text{t}-4)(\text{t}+3)}=\frac{\text{A}}{\text{t}-4}+\frac{\text{B}}{\text{t}+3}$
$\Rightarrow \text{t}=\text{A}(\text{t}+3)+\text{B}(\text{t}-4)$
On comparting the coefficient of t on both sides, we get
$\text{A}+\text{B}=1\ \ \dots(\text{i})$
Comparing constant term, we get
$3\text{A}-4\text{B}-0\ \ \dots(\text{ii})$
Solving (i) and (ii), we get
$\therefore\ \text{A}=\frac{4}{7}$ and $\text{B}=\frac{3}{7}$
$\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}=\frac{4}{7(\text{x}^2-4)}+\frac{3}{7(\text{x}^2+3)}$
$\therefore\ \text{I}=\frac{4}{7}\int\frac{1}{\text{x}^2-2^2}\text{dx}+\frac{3}{7}\int\frac{1}{\text{x}^2+(\sqrt{3})^2}\text{dx}$
$=\frac{4}{7}\cdot\frac{1}{2.2}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{3}{7}\cdot\frac{1}{\sqrt{3}}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
$=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$

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