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$\int \frac{1}{16 x^{2}+9} d x$ is equal to
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The correct answer is:
$\frac{1}{12} \tan ^{-1}\left(\frac{4 x}{3}\right)+c$
$\begin{aligned} \int \frac{1}{16 x^{2}+9} d x &=\frac{1}{16} \int \frac{1}{x^{2}+\left(\frac{3}{4}\right)^{2}} d x \\ &=\frac{1}{16} \times \frac{4}{3} \tan ^{-1}\left(\frac{x}{3 / 4}\right)+c \\ &=\frac{1}{12} \tan ^{-1}\left(\frac{4 x}{3}\right)+c \end{aligned}$
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