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

Sum

 ∫  {x-3} /{ x^2 + 2x - 4 } dx

#### Solution

$\int\left( \frac{x - 3}{x^2 + 2x - 4} \right)dx$
$x - 3 = A\frac{d}{dx}\left( x^2 + 2x - 4 \right) + B$
$x - 3 = A \left( 2x + 2 \right) + B$
$x - 3 = \left( 2 A \right) x + 2A + B$

Comparing Coefficients of like powers of x

$2A = 1$
$A = \frac{1}{2}$
$2A + B = - 3$
$2 \times \frac{1}{2} + B = - 3$
$B = - 4$

$Now, \int\left( \frac{x - 3}{x^2 + 2x - 4} \right)dx$
$= \int\left( \frac{\frac{1}{2}\left( 2x + 2 \right) - 4}{x^2 + 2x - 4} \right)dx$
$= \frac{1}{2}\int\frac{\left( 2x + 2 \right) dx}{\left( x^2 + 2x - 4 \right)} - 4\int\frac{dx}{x^2 + 2x + 1 - 1 - 4}$
$= \frac{1}{2}\int\frac{\left( 2x + 2 \right) dx}{\left( x^2 + 2x - 4 \right)} - 4\int\frac{dx}{\left( x + 1 \right)^2 - \left( \sqrt{5} \right)^2}$
$= \frac{1}{2} \text{ log }\left| x^2 + 2x - 4 \right| - \frac{4}{2\sqrt{5}} \text{ log }\left| \frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}} \right| + C$
$= \frac{1}{2} \text{ log }\left| x^2 + 2x - 4 \right| - \frac{2}{\sqrt{5}} \text{ log } \left| \frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}} \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.19 | Q 3 | Page 104