# ∫ X ( X 2 + 4 ) √ X 2 + 9 D X - Mathematics

Sum
$\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}} \text{ dx}$

#### Solution

$\text{ We have,}$
$I = \int\frac{x dx}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}}$
$\text{ Putting x}^2 = t$
$\Rightarrow 2x \text{ dx } = dt$
$\Rightarrow x \text{ dx} = \frac{dt}{2}$
$\therefore I = \frac{1}{2}\int\frac{dt}{\left( t + 4 \right) \sqrt{t + 9}}$
$\text{ Again Putting t} + 9 = u^2$
$\Rightarrow dt = 2u\text{ du }$
$\therefore I = \frac{1}{2}\int\frac{2u \text{ du}}{\left( u^2 - 9 + 4 \right) u}$
$= \int\frac{du}{u^2 - 5}$
$= \int\frac{du}{u^2 - \left( \sqrt{5} \right)^2}$
$= \frac{1}{2\sqrt{5}} \text{ log } \left| \frac{u - \sqrt{5}}{u + \sqrt{5}} \right| + C$
$= \frac{1}{2\sqrt{5}} \text{ log } \left| \frac{\sqrt{t + 9} - \sqrt{5}}{\sqrt{t + 9} + \sqrt{5}} \right| + C$
$= \frac{1}{2\sqrt{5}} \text{ log} \left| \frac{\sqrt{x^2 + 9} - \sqrt{5}}{\sqrt{x^2 + 9} + \sqrt{5}} \right| + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 14 | Page 196