Interesting Integrals 1

Substitute $\small{\color{teal}{ \ln x=t} \text{ such that } \color{teal}{\dfrac{\mathrm{d}x}{x}=\mathrm{d}t}}$ .

$\mathcal J=\int_2^{\infty}\dfrac{\mathrm{dt}}{t^3}=\left[\dfrac{-1}{2t^2}\right]_2^{\infty}=\left[\dfrac{1}{2t^2}\right]_{\infty}^2=\dfrac 18$

$\Large\therefore \sqrt{1+8}=\boxed{\color{#D61F06}{3}}$

$\begin{aligned} I & = \int_{e^2} ^\infty \frac 1{x\ln ^3 x} dx & \small \color{#3D99F6}{\text{Let }u = \ln x \implies e^u du = dx} \\ & = \int_2^\infty \frac {e^u}{e^u \cdot u^3} du \\ & = \int_2^\infty \frac 1{u^3} du \\ & = - \frac 1{2u^2} \bigg|_2^\infty \\ & = 0 + \frac 18 = \frac 18 \end{aligned}$

$\implies \sqrt{A+B} = \sqrt{1+8} = \boxed{3}$

@Krishna Shankar Pls post more of such problems! Looking forward to more! Thanks! :)