int 3^{-log _9 x^2} d x= | vedclass

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Question

(B) We are given the integral $I = \int 3^{-\log _9 x^2} d x$.Using the property of logarithms $n \log_b a = \log_b a^n$,we have $-\log _9 x^2 = \log

Vedclass · Accepted Answer

$\log |x|+C$

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(B) We are given the integral $I = \int 3^{-\log _9 x^2} d x$.Using the property of logarithms $n \log_b a = \log_b a^n$,we have $-\log _9 x^2 = \log _9 (x^2)^{-1} = \log _9 (\frac{1}{x^2})$.Substituting this into the integral,we get $I = \int 3^{\log _9 (\frac{1}{x^2})} d x$.Using the property $a^{\log _b c} = c^{\log _b a}$,we rewrite $3^{\log _9 (\frac{1}{x^2})}$ as $(\frac{1}{x^2})^{\log _9 3}$.Since $\log _9 3 = \log _{3^2} 3 = \frac{1}{2} \log _3 3 = \frac{1}{2}$,the expression becomes $(\frac{1}{x^2})^{\frac{1}{2}} = \frac{1}{x}$.Thus,$I = \int \frac{1}{x} d x = \log |x| + C$.

$\int 3^{-\log _9 x^2} d x=$

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