Question

Evaluate:

${\int }_1^e(\frac{1}{\sqrt{x\ln x}}+\sqrt{\frac{\ln x}{x}})dx$

Answer 1

Let $I={\int }_1^e(\frac{1}{\sqrt{xlnx}}+\sqrt{\frac{lnx}{x}})dx$

Take $x=e^y$, $\Rightarrow dx=e^{y}dy$

So, $x\to 1\Rightarrow e^y\to \ln 1$ and $x\to e\Rightarrow e^y\to \ln e$

Substituting in the integral and changing it's limits,

$\Rightarrow {\int }_{\ln 1}^{\ln e}(\frac{e^y}{\sqrt{y{e}^y}}+\sqrt{\frac{y}{e^y}}{e}^y)dy$

$=2{\int }_0^1{e}^{y/2}(\frac{1}{2}\frac{1}{\sqrt{y}}+\frac{1}{2}\sqrt{y})dy$

$=2\sqrt{y{e}^y}{]}_0^1=2\sqrt{e}$