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}}} $
$= \frac{1}{3} \sin^{-1} \left(\frac{3x+1}{2}\right)+c$