Question

$\int \frac{1}{\sqrt{3-6x-9x^2}}dx$ is equal to

• A. ${\sin }^{-1}(\frac{3x+1}{2})+C$
• B. ${\sin }^{-1}(\frac{3x+1}{6})+C$
• C. $\frac{1}{3}{\sin }^{-1}(\frac{3x+1}{2})+C$
• D. ${\sin }^{-1}(\frac{2x+1}{3})+C$

Answer 1

The correct option is C $\frac{1}{3}{\sin }^{-1}(\frac{3x+1}{2})+C$

$\int \frac{1}{\sqrt{3-6x-9x^2}}dx$
$=\int \frac{1}{\sqrt{4-(3x+1{)}^2}}dx$
$=\frac{1}{3}{\sin }^{-1}(\frac{3x+1}{2})+C$