evaluate the definite integral $int_{3}^{9}\frac{1}{x(ln(x))^{2}}dx$\n$int_{3}^{9}\frac{1}{x(ln(x))^{2}}dx =…

Explanation:

Step1: Use substitution

Let $u = \ln(x)$, then $du=\frac{1}{x}dx$. When $x = 3$, $u=\ln(3)$; when $x = 9$, $u=\ln(9)$.

Step2: Rewrite the integral

The integral $\int_{3}^{9}\frac{1}{x(\ln(x))^{2}}dx$ becomes $\int_{\ln(3)}^{\ln(9)}\frac{1}{u^{2}}du$.

Step3: Integrate $\frac{1}{u^{2}}$

We know that $\int u^{- 2}du=-\frac{1}{u}+C$. So, $\int_{\ln(3)}^{\ln(9)}\frac{1}{u^{2}}du=\left[-\frac{1}{u}\right]_{\ln(3)}^{\ln(9)}$.

Step4: Evaluate the definite - integral

$-\frac{1}{\ln(9)}+\frac{1}{\ln(3)}=-\frac{1}{2\ln(3)}+\frac{1}{\ln(3)}=\frac{- 1 + 2}{2\ln(3)}=\frac{1}{2\ln(3)}$.

Answer:

$\frac{1}{2\ln(3)}$