int frac{1}{sqrt{3-6 x-9 x^{2}}} d x is equ | vedclass

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(C) To evaluate the integral $ I = \int \frac{1}{\sqrt{3-6 x-9 x^{2}}} d x $,we first complete the square for the quadratic expression inside the squa

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) To evaluate the integral $ I = \int \frac{1}{\sqrt{3-6 x-9 x^{2}}} d x $,we first complete the square for the quadratic expression inside the square root. $ 3-6 x-9 x^{2} = 3 - (9 x^{2} + 6 x) = 3 - ((3 x)^{2} + 2(3 x)(1) + 1^{2} - 1) = 3 - ((3 x+1)^{2} - 1) = 4 - (3 x+1)^{2} $. Now,the integral becomes $ I = \int \frac{1}{\sqrt{4-(3 x+1)^{2}}} d x $. Let $ u = 3x+1 $,then $ du = 3 dx $,which implies $ dx = \frac{du}{3} $. Substituting these into the integral,we get $ I = \int \frac{1}{\sqrt{2^{2}-u^{2}}} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{1}{\sqrt{2^{2}-u^{2}}} du $. Using the standard formula $ \int \frac{1}{\sqrt{a^{2}-x^{2}}} dx = \sin^{-1}(\frac{x}{a}) + C $,we obtain $ I = \frac{1}{3} \sin^{-1}(\frac{u}{2}) + C $. Substituting $ u = 3x+1 $ back,we get $ I = \frac{1}{3} \sin^{-1}(\frac{3x+1}{2}) + C $.

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

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