# ∫ X 2 + 1 X 2 − 5 X + 6 D X - Mathematics

Sum
$\int\frac{x^2 + 1}{x^2 - 5x + 6} \text{ dx }$

#### Solution

$\text{ Let }I\int\left( \frac{x^2 + 1}{x^2 - 5x + 6} \right)dx$
$\text{Dividing Numerator by Denominator}$

$\frac{x^2 + 1}{x^2 - 5x + 6} = 1 + \left( \frac{5x - 5}{x^2 - 5x + 6} \right) . . . . . \left( 1 \right)$
$\text{ Also } \frac{5x - 5}{x^2 - 5x + 6} = \frac{5x - 5}{\left( x - 2 \right) \left( x - 3 \right)}$
$\text{ Let } \frac{5x - 5}{\left( x - 2 \right) \left( x - 3 \right)} = \frac{A}{x - 2} + \frac{B}{x - 3}$
$\Rightarrow \frac{5x - 5}{\left( x - 2 \right) \left( x - 3 \right)} = \frac{A \left( x - 3 \right) + B \left( x - 2 \right)}{\left( x - 2 \right) \left( x - 3 \right)}$
$\Rightarrow 5x - 5 = A \left( x - 3 \right) + B \left( x - 2 \right)$
$\text{ let } x = 3$
$5 \times 3 - 5 = A \times 0 + B \left( 3 - 2 \right)$
$10 = B$
$\text{ let } x = 2$
$5 \times 2 - 5 = A \left( 2 - 3 \right) + B \times 0$
$A = - 5$
$\frac{5x - 5}{\left( x - 2 \right) \left( x - 3 \right)} = \frac{- 5}{x - 2} + \frac{10}{x - 3} . . . . . \left( 2 \right)$
$\text{ from }\left( 1 \right) \text{ and }\left( 2 \right)$
$I = \int dx - 5\int\frac{dx}{x - 2} + 10\int\frac{dx}{x - 3}$
$= x - 5 \text{ log } \left| x - 2 \right| + 10 \text{ log } \left| x - 3 \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.2 | Q 4 | Page 106