Mathematics

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


ANSWER

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


SOLUTION
$$3-6x-9x^2=4-(9x^2+6x+1)=4-(3x+1)^2$$

$$\displaystyle \therefore\int\dfrac{1}{\sqrt{4-(3x+1)^2}}dx=\dfrac{1}{3}\sin^{-1}\left(\dfrac{3x+1}{2}\right)+c$$.
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Single Correct Medium Published on 17th 09, 2020
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