Question

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

• A. $-\frac{2}{3}\sqrt{3\left(\frac{1}{x+1}\right)-1}+C$
• B. $\frac{2}{3}\sqrt{3\left(\frac{1}{x-1}\right)-1}+C$
• C. $\frac{3}{2}\sqrt{3\left(\frac{1}{x-1}\right)-1}+C$
• D. $-\frac{4}{3}\sqrt{3\left(\frac{1}{x-1}\right)-1}+C$

Answer 1

The correct option is A $-\frac{2}{3}\sqrt{3\left(\frac{1}{x+1}\right)-1}+C$

$I=\int \frac{dx}{(1+x)\sqrt{2+x-x^2}}$
$=\int \frac{dx}{(1+x)\sqrt{(1+x)(2-x)}}$
Put $x+1=\frac{1}{t}$
$\Rightarrow dx=-\frac{1}{t^2}dt$
$\therefore I=-\int \frac{dt}{\sqrt{3t-1}}$
Put $3t-1=u$
$\Rightarrow dt=\frac{du}{3}$
$\therefore I=-\frac{1}{3}\int \frac{du}{\sqrt{u}}$
$=-\frac{2}{3}\sqrt{u}+C$
$=-\frac{2}{3}\sqrt{3t-1}+C$
$I=-\frac{2}{3}\sqrt{3\left(\frac{1}{x+1}\right)-1}+C$