Q118 / 180 · KCET 2018 · Mathematics
\( \int \frac{1}{\sqrt{3-6 x-9 x^{2}}} d x \) is equal to
- A\( \sin ^{-1}\left(\frac{3 x+1}{2}\right)+C \)
- B\( \sin ^{-1}\left(\frac{3 x+1}{6}\right)+C \)
- Correct answerC\( \frac{1}{3} \sin ^{-1}\left(\frac{3 x+1}{2}\right)+C \)
- D\( \sin ^{-1}\left(\frac{2 x+1}{3}\right)+C \)
Solution
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, \( \int \frac{1}{\sqrt{4-(3 x+1)^{2}}} d x \) \( =\frac{1}{3} \sin ^{-1}\left(\frac{3 x+1}{2}\right)+c \)
Since, \( 3-6 x-9 x^{2}=4-(3 x+1)^{2} \)
So, \( \int \frac{1}{\sqrt{4-(3 x+1)^{2}}} d x \) \( =\frac{1}{3} \sin ^{-1}\left(\frac{3 x+1}{2}\right)+c \)
Hence, the correct answer is option: 'C'.