# ∫ X 2 − 2 X 5 − X Dx - Mathematics

Sum
$\int\frac{x^2 - 2}{x^5 - x} \text{ dx}$

#### Solution

$\text{We have},$
$I = \int\left( \frac{x^2 - 2}{x^5 - x} \right) dx$
$= \int\frac{\left( x^2 - 2 \right)}{x \left( x^4 - 1 \right)}dx$
$= \int\frac{x \left( x^2 - 2 \right)}{x^2 \left( x^2 - 1 \right) \left( x^2 + 1 \right)}dx$
$\text{ Putting x^2 = t}$
$\Rightarrow 2x\ dx = dt$
$\Rightarrow x\ dx = \frac{dt}{2}$
$\therefore I = \frac{1}{2}\int\frac{\left( t - 2 \right)}{t \left( t - 1 \right) \left( t + 1 \right)}\text{ dt }$
$\text{ Let } \frac{t - 2}{t \left( t - 1 \right) \left( t + 1 \right)} = \frac{A}{t} + \frac{B}{t - 1} + \frac{C}{t + 1}$
$\Rightarrow \frac{t - 2}{t \left( t - 1 \right) \left( t + 1 \right)} = \frac{A \left( t - 1 \right) \left( t + 1 \right) + Bt \left( t + 1 \right) + Ct \cdot \left( t - 1 \right)}{t \left( t - 1 \right) \left( t + 1 \right)}$
$\Rightarrow t - 2 = A \left( t - 1 \right) \left( t + 1 \right) + B t \left( t + 1 \right) + C t \left( t - 1 \right)$
$\text{ Putting t = 1}$
$\therefore 1 - 2 = B \times 2$
$\Rightarrow B = - \frac{1}{2}$
$\text{ Putting t = 0}$
$\therefore - 2 = A \left( - 1 \right)$
$\Rightarrow A = 2$
$\text{ Putting t = - 1}$
$\therefore - 3 = C \left( - 1 \right) \left( - 2 \right)$
$\Rightarrow C = - \frac{3}{2}$
$\therefore I = \frac{2}{2}\int\frac{dt}{t} - \frac{1}{2 \times 2}\int\frac{dt}{t - 1} - \frac{3}{2 \times 2}\int\frac{d}{t + 1}$
$= \text{ log} \left| t \right| - \frac{1}{4} \text{ log }\left| t - 1 \right| - \frac{3}{4} \text{ log} \left| t + 1 \right| + C$
$= \text{ log }\left| x^2 \right| - \frac{1}{4} \text{ log }\left| x^2 - 1 \right| - \frac{3}{4} \text{ log} \left| x^2 + 1 \right| + C$
$= 2 \text{ log } \left| x \right| - \frac{1}{4} \text{ log} \left| x^2 - 1 \right| - \frac{3}{4} \text{ log} \left| x^2 + 1 \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Revision Excercise | Q 126 | Page 205