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# ∫ X 2 X 2 + 7 X + 10 D X - Mathematics

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Sum
$\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }$
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#### Solution

$Let\text{ I } = \int\left( \frac{x^2}{x^2 + 7x + 10} \right)dx$
$\text{ Now },$

$x^2 + 7x + 10 {x^2}^1$
$x^2 + 7x + 10$
$- - -$
$- 7x - 10$
$\therefore \frac{x^2}{x^2 + 7x + 10} = 1 - \frac{\left( 7x + 10 \right)}{x^2 + 7x + 10}$
$\Rightarrow \frac{x^2}{x^2 + 7x + 10} = 1 - \left( \frac{7x + 10}{x^2 + 2x + 5x + 10} \right)$
$\frac{x^2}{x^2 + 7x + 10} = 1 - \left( \frac{7x + 10}{x \left( x + 2 \right) + 5 \left( x + 2 \right)} \right)$
$\frac{x^2}{x^2 + 7x + 10} = 1 - \left[ \frac{7x + 10}{\left( x + 2 \right) \left( x + 5 \right)} \right] . . . . . \left( 1 \right)$
$\text{ Consider },$
$\frac{7x + 10}{\left( x + 2 \right) \left( x + 5 \right)} = \frac{A}{\left( x + 2 \right)} + \frac{B}{x + 5}$
$7x + 10 = A \left( x + 5 \right) + B \left( x + 2 \right)$
$\text{ let } x + 5 = 0$
$x = - 5$
$\Rightarrow 7 \left( - 5 \right) + 10 = A \times 0 + B \left( - 5 + 2 \right)$
$- 25 = B \left( - 3 \right)$
$\Rightarrow B = \frac{25}{3}$
$\text{ let } x + 2 = 0$
$x = - 2$
$7 \left( - 2 \right) + 10 = A \left( - 2 + 5 \right)$
$\Rightarrow - 4 = A \left( 3 \right)$
$\Rightarrow A = - \frac{4}{3}$
$\frac{7x + 10}{\left( x + 2 \right) \left( x + 5 \right)} = \frac{- 4}{3 \left( x + 2 \right)} + \frac{25}{3 \left( x + 5 \right)} . . . . . \left( 2 \right)$
$\text{ from } \left( 1 \right) \text { and } \left( 2 \right)$
$\frac{x^2}{x^2 + 7x + 10} = 1 + \frac{4}{3 \left( x + 2 \right)} - \frac{25}{3 \left( x + 5 \right)}$
$\Rightarrow \int\frac{x^2 dx}{x^2 + 7x + 10} = \int dx + \frac{4}{3}\int\frac{dx}{x + 2} - \frac{25}{3}\int\frac{dx}{x + 5}$
$= x + \frac{4}{3} \text{ log } \left| x + 2 \right| - \frac{25}{3} \text{ log } \left| x + 5 \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.2 | Q 5 | Page 106

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