# Evaluate the Following Integral: ∫ X 2 X 4 − X 2 − 12 D X - Mathematics

Sum

Evaluate the following integral:

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

#### Solution

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

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

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

Concept: Evaluation of Simple Integrals of the Following Types and Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 66 | Page 178