KCET · Maths · Indefinite Integration
\( \int \frac{1}{\sqrt{3-6 x-9 x^{2}}} d x \) is equal to
- A \( \sin ^{-1}\left(\frac{3 x+1}{2}\right)+C \)
- B \( \sin ^{-1}\left(\frac{3 x+1}{6}\right)+C \)
- C \( \frac{1}{3} \sin ^{-1}\left(\frac{3 x+1}{2}\right)+C \)
- D \( \sin ^{-1}\left(\frac{2 x+1}{3}\right)+C \)
Answer & Solution
Correct Answer
(C) \( \frac{1}{3} \sin ^{-1}\left(\frac{3 x+1}{2}\right)+C \)
Step-by-step Solution
Detailed explanation
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, \( \int \frac{1}{\sqrt{4-(3 x+1)^{2}}} d x \) \( =\frac{1}{3} \sin ^{-1}\left(\frac{3 x+1}{2}\right)+c \)
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