# ∫ 2 X + 5 X 2 − X − 2 D X - Mathematics

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

#### Solution

$\int\frac{\left( 2x + 5 \right) dx}{x^2 - x - 2}$
$2x + 5 = A\frac{d}{dx}\left( x^2 - x - 2 \right) + B$
$2x + 5 = A \left( 2x - 1 \right) + B$
$2x + 5 = \left( 2 A \right) x + B - A$

Comparing the Coefficients of like powers of x

$2 A = 2$
$A = 1$
$B - A = 5$
$B - 1 = 5$
$B = 6$

$\therefore 2x + 5 = 1 \cdot \left( 2x - 1 \right) + 6$
$\therefore \int\left( \frac{2x + 5}{x^2 - x - 2} \right)dx$
$\Rightarrow \int\left( \frac{\left( 2x - 1 \right) + 6}{x^2 - x - 2} \right)dx$
$\Rightarrow \int\left( \frac{2x - 1}{x^2 - x - 2} \right)dx + 6\int\frac{dx}{x^2 - x - 2}$
$= I_1 + 6 I_2 \left( \text{ say }\right) . . . \left( 1 \right)$
$\text{ where }$
$I_1 = \int\left( \frac{2x - 1}{x^2 - x - 2} \right)\text{ dx }I_2 = \int\frac{dx}{x^2 - x - 2}$
$I_1 = \int\left( \frac{2x - 1}{x^2 - x - 2} \right)dx$
$\text{ let x}^2 - x - 2 = t$
$\Rightarrow \left( 2x - 1 \right) dx = dt$
$I_1 = \int\frac{dt}{t}$
$I_1 = \text{ log }\left( t \right)$
$I_1 = \text{ log }\left| x^2 - x - 2 \right| + C_1 . . . \left( 2 \right)$
$I_2 = \int\frac{dx}{x^2 - x - 2}$
$I_2 = \int\frac{dx}{x^2 - x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 - 2}$
$I_2 = \int\frac{dx}{\left( x - \frac{1}{2} \right)^2 - \frac{1}{4} - 2}$
$I_2 = \int\frac{dx}{\left( x - \frac{1}{2} \right)^2 - \left( \frac{3}{2} \right)^2}$
$I_2 = \frac{1}{2 \times \frac{3}{2}} \text{ log }\left| \frac{x - \frac{1}{2} - \frac{3}{2}}{x - \frac{1}{2} + \frac{3}{2}} \right|$
$I_2 = \frac{1}{3} \text{ log} \left| \frac{x - 2}{x + 1} \right| + C_2 . . . \left( 3 \right)$
$\int\frac{\left( 2x + 5 \right) dx}{x^2 - x - 2}$
$= \text{ log} \left| x^2 - x - 2 \right| + \frac{6}{3} \text{ log} \left| \frac{x - 2}{x + 1} \right| + C_1 + C_2$
$= \text{ log }\left| x^2 - x - 2 \right| + 2 \text{ log }\left| \frac{x - 2}{x + 1} \right| + C \left( \text{ Where C }= C_1 + C_2 \right)$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.19 | Q 8 | Page 104