# ∫ 2 X 2 + 7 X − 3 X 2 ( 2 X + 1 ) D X - Mathematics

Sum
$\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx$

#### Solution

We have,

$I = \int\frac{\left( 2 x^2 + 7x - 3 \right) dx}{x^2 \left( 2x + 1 \right)}$

$\text{Let }\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x + 1}$

$\Rightarrow \frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} = \frac{A \left( x \right) \left( 2x + 1 \right) + B \left( 2x + 1 \right) + C x^2}{x^2 \left( 2x + 1 \right)}$

$\Rightarrow 2 x^2 + 7x - 3 = A \left( 2 x^2 + x \right) + B \left( 2x + 1 \right) + C x^2$

$\Rightarrow 2 x^2 + 7x - 3 = \left( 2A + C \right) x^2 + \left( A + 2B \right)x + B$

$\text{Equating coefficients of like terms}$

$2A + C = 2 . . . . . \left( 1 \right)$

$A + 2B = 7 . . . . . \left( 2 \right)$

$B = - 3 . . . . . \left( 3 \right)$

$\text{Solving (1), (2) and (3), we get}$

$A = 13$

$B = - 3$

$C = - 24$

$\therefore \frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} = \frac{13}{x} - \frac{3}{x^2} - \frac{24}{2x + 1}$

$\Rightarrow I = 13\int\frac{dx}{x} - 3\int x^{- 2} dx - 24\int\frac{dx}{2x + 1}$

$= 13 \log \left| x \right| + \frac{3}{x} - 24 \frac{\log \left| 2x + 1 \right|}{2} + C$

$= 13 \log \left| x \right| + \frac{3}{x} - 12 \log \left| 2x + 1 \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 33 | Page 177