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

Question

$ \int \frac{1}{\sqrt{3-6 x-9 x^{2}}} d x $ is equal to, $ \sin ^{-1}\left(\frac{3 x+1}{2}\right)+C $, $ \sin ^{-1}\left(\frac{3 x+1}{6}\right)+C $, $ \frac{1}{3} \sin ^{-1}\left(\frac{3 x+1}{2}\right)+C $, $ \sin ^{-1}\left(\frac{2 x+1}{3}\right)+C $

MARKS App · Accepted Answer

Given that, $ \int \frac{1}{\sqrt{3-6 x-9 x^{2}}} d x $ Since, $ 3-6 x-9 x^{2}=4-(3 x+1)^{2} $ So, $ \int \frac{1}{\sqrt{4-(3 x+1)^{2}}} d x $ $ =\frac{1}{3} \sin ^{-1}\left(\frac{3 x+1}{2}\right)+c $

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