Question

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

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

Answer 1

Answer: (C)

$\int \frac{1}{\sqrt{3-6x-9x^2}}dx=\int \frac{dx}{\sqrt{-\left[9x^2+6x-3\right]}}$
$=\int \frac{dx}{\sqrt{-\left[9x^2+6x+1-4\right]}}$
$=\int \frac{dx}{\sqrt{-\left[{\left(3x+1\right)}^2-2^2\right]}}$
$=\int \frac{dx}{\sqrt{2^2-{\left(3x+1\right)}^2}}$
$=\frac{1}{3}{\sin }^{-1}\left(\frac{3x+1}{2}\right)+c$