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$\int \frac{d x}{\sqrt{7-6 x-x^2}}$ is equal to
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The correct answer is:
$\sin ^{-1}\left(\frac{x+3}{4}\right)+c$
$I=\int \frac{d x}{\sqrt{7-6 x-x^2}}$
$\begin{aligned} I & =\int \frac{d x}{\sqrt{7-6 x-x^2+9-9}} \\ & =\int \frac{d x}{\sqrt{(4)^2-(x+3)^2}}\left\{\because \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1}\left(\frac{x}{a}\right)+C\right. \\ I & =\sin ^{-1}\left(\frac{x+3}{4}\right)+C\end{aligned}$
$\begin{aligned} I & =\int \frac{d x}{\sqrt{7-6 x-x^2+9-9}} \\ & =\int \frac{d x}{\sqrt{(4)^2-(x+3)^2}}\left\{\because \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1}\left(\frac{x}{a}\right)+C\right. \\ I & =\sin ^{-1}\left(\frac{x+3}{4}\right)+C\end{aligned}$
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