Answer
$$\int^{e^2}_e\frac{1}{x\sqrt{\ln x}}dx=2(\sqrt2-1)$$
Work Step by Step
$$A=\int^{e^2}_e\frac{1}{x\sqrt{\ln x}}dx$$ $$A=\int^{e^2}_e\frac{(\ln x)^{-1/2}}{x}dx$$
Set $u=\ln x$, which means $$du=\frac{1}{x}dx$$
- For $x=e$, we have $u=\ln e=1$
- For $x=e^2$, we have $u=\ln e^2=2$
Therefore, $$A=\int^2_1 u^{-1/2}du$$ $$A=\frac{u^{1/2}}{\frac{1}{2}}\Big]^2_1=2\sqrt u\Big]^2_1$$ $$A=2(\sqrt2-1)$$
