# ∫ X 2 + 6 X − 8 X 3 − 4 X D X - Mathematics

Sum
$\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx$

#### Solution

We have,

$I = \int\left( \frac{x^2 + 6x - 8}{x^3 - 4x} \right)dx$

$= \int\frac{\left( x^2 + 6x - 8 \right)}{x \left( x^2 - 4 \right)}dx$

$= \int\frac{\left( x^2 + 6x - 8 \right)}{x \left( x - 2 \right) \left( x + 2 \right)}dx$

$Let \frac{x^2 + 6x - 8}{x \left( x - 2 \right) \left( x + 2 \right)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}$

$\Rightarrow \frac{x^2 + 6x - 8}{x \left( x - 2 \right) \left( x + 2 \right)} = \frac{A \left( x - 2 \right) \left( x + 2 \right) + B \left( x \right) \left( x + 2 \right) + C \left( x \right) \left( x - 2 \right)}{x \left( x - 2 \right) \left( x + 2 \right)}$

$\Rightarrow x^2 + 6x - 8 = A \left( x^2 - 4 \right) + B \left( x^2 + 2x \right) + C \left( x^2 - 2x \right)$

Putting x - 2 = 0

$\Rightarrow x = 2$

$4 + 6 \times 2 - 8 = A \times 0 + B \left( 4 + 4 \right)$

$\Rightarrow 8 = B \times 8$

$\Rightarrow B = 1$

Putting x = - 2

$4 - 12 - 8 = A \times 0 + B \times 0 + C \times 8$
$\Rightarrow C = - 2$

Putting x = 0

$- 8 = A \left( - 4 \right) + B \times 0 + C \times 0$

$\Rightarrow A = 2$

$\therefore I = \int\frac{2}{x} + \int\frac{dx}{x - 2} - 2\int\frac{dx}{x + 2}$

$= 2 \log \left| x \right| + \log \left| x - 2 \right| - 2 \log \left| x + 2 \right| + C$

$= \log x^2 + \log \left| x - 2 \right| - \log \left| x + 2 \right|^2 + C$

$= \log \left| \frac{x^2 \left( x - 2 \right)}{\left( x + 2 \right)^2} \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 20 | Page 176