$\int \frac{x}{(x^{2} + 2 x + 2) \sqrt{x + 1}} d x is equal to$

Moderate

A

$\frac{- 1}{\sqrt{2}} \log |\frac{x + 1 + \sqrt{2} \sqrt{x + 1} + 1}{x + 1 - \sqrt{2} \sqrt{x + 1} + 1}| + C$
B

$\frac{1}{\sqrt{2}} \log |\frac{x + 1 - \sqrt{2} \sqrt{x + 1} + 1}{x + 1 + \sqrt{2} \sqrt{x + 1} + 1}| + C$
C

$\frac{1}{\sqrt{2}} \log |\frac{x + 1 + \sqrt{2} \sqrt{x + 1} + 1}{x + 1 - \sqrt{2} \sqrt{x + 1} + 1}| + C$
D

$\frac{- 1}{\sqrt{2}} \log |\frac{x + 1 - \sqrt{2} \sqrt{x + 1} + 1}{x + 1 + \sqrt{2} \sqrt{x + 1} + 1}| + C$

Solution

$\int \frac{x}{(x^{2} + 2 x + 2) \sqrt{x + 1}} d x = \int \frac{x}{[{(x + 1)}^{2} + 1] \sqrt{x + 1}} d x$
$Put x + 1 = t^{2}$
$\begin{array}{l} d x = 2 t d t \\ \int \frac{(t^{2} - 1) 2 t d t}{(t^{4} + 1) t} \\ = 2 \int \frac{t^{2} (1 - \frac{1}{t^{2}}) d t}{t^{2} (t^{2} + \frac{1}{t^{2}})} \\ = 2 \int \frac{1 - \frac{1}{t^{2}}}{(t^{2} + \frac{1}{t^{2}})} d t \\ = 2 \int \frac{(1 - \frac{1}{t^{2}}) d t}{{(t + \frac{1}{t})}^{2} - (\sqrt{2})^{2}} \end{array}$
$Put t + \frac{1}{t} = k, (1 - \frac{1}{t^{2}}) d t = d k$
$\begin{array}{l} = 2 \int \frac{d k}{k^{2} - (\sqrt{2})^{2}} \\ = 2 \frac{1}{2 \sqrt{2}} \log |\frac{k - \sqrt{2}}{k + \sqrt{2}}| + C \\ = \frac{1}{\sqrt{2}} \log |\frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}}| + C \\ = \frac{1}{\sqrt{2}} \log |\frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1}| + C \end{array}$
$= \frac{1}{\sqrt{2}} \log |\frac{x + 1 - \sqrt{2} \sqrt{x + 1} + 1}{x + 1 + \sqrt{2} \sqrt{x + 1} + 1}| + C$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME