Question

$\int \sqrt{1-4x-x^2}dx$ is equal to
$($where $C$ is integration constant$)$

• A. $\frac{x+2}{2}\sqrt{1-4x-x^2}+\frac{5}{2}{\sin }^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C$
• B. $\frac{x}{2}\sqrt{1-4x-x^2}+\frac{5}{2}{\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)+C$
• C. $\frac{5}{2}{\sin }^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C$
• D. $\frac{x}{2}\sqrt{1-4x-x^2}+\frac{5}{2}\ln |x+2+\sqrt{1-4x-x^2}|+C$

Answer 1

The correct option is A $\frac{x+2}{2}\sqrt{1-4x-x^2}+\frac{5}{2}{\sin }^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C$

$I=\int \sqrt{1-4x-x^2}dx=\int \sqrt{{\left(\sqrt{5}\right)}^2-(x+2{)}^2}dx\therefore I=\frac{x+2}{2}\sqrt{1-4x-x^2}+\frac{5}{2}{\sin }^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C[\because \int \sqrt{a^2-x^2}dx=\frac{x}{a}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C]$