Question

$\int \frac{dx}{(x+1)\sqrt{1+x-x^2}}$ equals to (where $C$ is integration constant)

• A. ${\sin }^{-1}\left[\frac{3x+1}{x+1}\right]+C$
• B. $\frac{1}{\sqrt{5}}{\sin }^{-1}\left[\frac{3x+1}{\sqrt{5}(x+1)}\right]+C$
• C. ${\sin }^{-1}\left[\frac{3x+1}{\sqrt{5}(x+1)}\right]+C$
• D. $\frac{1}{2\sqrt{5}}{\sin }^{-1}\left[\frac{3x+1}{x+1}\right]+C$

Answer 1

The correct option is C ${\sin }^{-1}\left[\frac{3x+1}{\sqrt{5}(x+1)}\right]+C$

$I=\int \frac{dx}{(x+1)\sqrt{1+x-x^2}}$
Put $x+1=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt$ $\Rightarrow I=\int \frac{-\frac{1}{t^2}dt}{\frac{1}{t}\sqrt{\frac{1}{t}-{(\frac{1}{t}-1)}^2}}=\int \frac{-dt}{\sqrt{3t-1-t^2}}$
$=\int \frac{-dt}{\sqrt{\frac{5}{4}-{(t-\frac{3}{2})}^2}}=-{\sin }^{-1}\left(\frac{2t-3}{\sqrt{5}}\right)+C$
$=-{\sin }^{-1}\left(\frac{\frac{2}{x+1}-3}{\sqrt{5}}\right)+C$
$={\sin }^{-1}\left(\frac{3x+1}{\sqrt{5}(x+1)}\right)+C$