∫ X ( X 2 + 2 X + 2 ) √ X + 1 D X - Mathematics

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

Solution

$\text{ We have,}$
$I = \int \frac{x \text{ dx}}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}}$
$= \int \frac{x \text{ dx}}{\left[ \left( x + 1 \right)^2 + 1 \right] \sqrt{x + 1}}$
$\text{ Putting x }+ 1 = t^2$
$\Rightarrow x = t^2 - 1$
$\text{ Diff both sides}$
$dx = 2t \text{ dt}$
$\therefore I = \int \frac{\left( t^2 - 1 \right)2t dt}{\left[ \left( t^2 \right)^2 + 1 \right] t}$
$= 2\int \frac{\left( t^2 - 1 \right)dt}{t^4 + 1}$
$\text{Dividing numerator and denominator by} \text{ t}^2$
$I = 2\left( \frac{1 - \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \right)dt$

$= 2\int\frac{\left( 1 - \frac{1}{t^2} \right)dt}{t^2 + \frac{1}{t^2} + 2 - 2}$
$= 2\int \frac{\left( 1 - \frac{1}{t^2} \right)dt}{\left( t + \frac{1}{t} \right)^2 - \left( \sqrt{2} \right)^2}$
$\text{ Putting t }+ \frac{1}{t} = p$
$\Rightarrow \left( 1 - \frac{1}{t^2} \right)dt = dp$
$I = 2\int \frac{dp}{p^2 - \left( \sqrt{2} \right)^2}$
$= 2 \times \frac{1}{2\sqrt{2}}\text{ log} \left| \frac{p - \sqrt{2}}{p + \sqrt{2}} \right| + C$
$= \frac{1}{\sqrt{2}}\text{ log }\left| \frac{p - \sqrt{2}}{P + \sqrt{2}} \right| + C$
$= \frac{1}{\sqrt{2}}\text{ log} \left| \frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}} \right| + C$
$= \frac{1}{\sqrt{2}}\text{ log} \left| \frac{t^2 - \sqrt{2}t + 1}{t^2 + \sqrt{2}t + 1} \right| + C$
$= \frac{1}{\sqrt{2}}\text{ log }\left| \frac{x + 1 - \sqrt{2\left( x + 1 \right)} + 1}{x + 1 + \sqrt{2\left( x + 1 \right)} + 1} \right| + C$
$= \frac{1}{\sqrt{2}}\text{ log} \left| \frac{\left( x + 2 \right) - \sqrt{2\left( x + 1 \right)}}{\left( x + 2 \right) + \sqrt{2\left( x + 1 \right)}} \right| + C$

Concept: Indefinite Integral Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 7 | Page 196