evaluate the definite integral.\n int_{e}^{e^{36}} \frac{1}{xsqrt{ln(x)}} dx

Explanation:

Step1: Use substitution

Let $u = \ln(x)$. Then $du=\frac{1}{x}dx$. When $x = e$, $u=\ln(e) = 1$; when $x=e^{36}$, $u=\ln(e^{36})=36$.

Step2: Rewrite the integral

The integral $\int_{e}^{e^{36}}\frac{1}{x\sqrt{\ln(x)}}dx$ becomes $\int_{1}^{36}\frac{1}{\sqrt{u}}du$.

Step3: Rewrite the integrand

We can rewrite $\frac{1}{\sqrt{u}}$ as $u^{-\frac{1}{2}}$.

Step4: Integrate

Using the power - rule for integration $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n=-\frac{1}{2}$, we have $\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C = 2u^{\frac{1}{2}}+C$.

Step5: Evaluate the definite integral

$2u^{\frac{1}{2}}\big|_{1}^{36}=2\sqrt{36}-2\sqrt{1}$. $2\times6 - 2\times1=12 - 2$.

Answer:

$10$