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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

$\log |x|+C$

Vedclass · Answer

(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,$I = \int \frac{1}{x} d x = \log |x| + C$.

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

Similar Questions

$\int \frac{dx}{\cos 2x - \cos^2 x} = $

Integrate the function: $\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}$

$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$ is equal to

Let $f(x)$ be an indefinite integral of $\cos^3 x$.
Statement $1$: $f(x)$ is a periodic function of period $\pi$.
Statement $2$: $\cos^3 x$ is a periodic function.

Find the integral of the function $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module