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$\int 3^{-\log _9 x^2} d x=$
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Verified Answer
The correct answer is:
$\log |x|+C$
$\int 3^{-\log _9 x^2} d x=\int 3^{\log _9\left(\frac{1}{x^2}\right)} d x$
$=\int\left(\frac{1}{x^2}\right)^{\log _9 3} d x=\int\left(\frac{1}{x^2}\right)^{\frac{1}{2}} d x$
$=\int \frac{1}{x} d x=\log |x|+C$.
$=\int\left(\frac{1}{x^2}\right)^{\log _9 3} d x=\int\left(\frac{1}{x^2}\right)^{\frac{1}{2}} d x$
$=\int \frac{1}{x} d x=\log |x|+C$.
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