∫ 1 ( X 2 + 2 ) ( X 2 + 5 ) Dx - Mathematics

Sum
$\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}$

Solution

$\text{We have},$
$I = \int\frac{dx}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)}$
$\text{ Putting x}^2 = t$
$\therefore \frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} = \frac{1}{\left( t + 2 \right) \left( t + 5 \right)}$
$\text{ Let }\frac{1}{\left( t + 2 \right) \left( t + 5 \right)} = \frac{A}{t + 2} + \frac{B}{t + 5}$
$\Rightarrow \frac{1}{\left( t + 2 \right) \left( t + 5 \right)} = \frac{A \left( t + 5 \right) + B \left( t + 2 \right)}{\left( t + 2 \right) \left( t + 5 \right)}$
$\Rightarrow 1 = A \left( t + 5 \right) + B \left( t + 2 \right)$
$\text{ Putting t = - 5}$
$\therefore 1 = B \left( - 5 + 2 \right)$
$\Rightarrow B = - \frac{1}{3}$
$\text{ Putting t = - 2}$
$\therefore 1 = A \left( - 2 + 5 \right) + B \times 0$
$\Rightarrow A = \frac{1}{3}$
$\therefore I = \frac{1}{3}\int\frac{dx}{x^2 + 2} - \frac{1}{3}\int\frac{dx}{x^2 + 5}$
$= \frac{1}{3}\int\frac{dx}{x^2 + \left( \sqrt{2} \right)^2} - \frac{1}{3}\int\frac{dx}{x^2 + \left( \sqrt{5} \right)^2}$
$= \frac{1}{3\sqrt{2}} \text{ tan}^{- 1} \left( \frac{x}{\sqrt{2}} \right) - \frac{1}{3\sqrt{5}} \text{ tan}^{- 1} \left( \frac{x}{\sqrt{5}} \right) + C$

Concept: Indefinite Integral Problems
Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Revision Excercise | Q 125 | Page 205