Calculus For All

Presenting @Ayush Sharma 's solution in LaTex.

$\begin{aligned} I & = \int \frac {x-1}{(x+x\sqrt x+\sqrt x)\sqrt{\sqrt x(x+1)}} dx & \small \color{#3D99F6} \text{Divide up and down by }x\sqrt x \\ & = \int \frac {\frac 1{\sqrt x}-\frac 1{x\sqrt x}}{\frac {x+x\sqrt x+\sqrt x}x\sqrt{\frac {\sqrt x(x+1)}x}} dx \\ & = \int \frac {\frac 1{\sqrt x}-\frac 1{x\sqrt x}}{\left(1+\sqrt x+ \frac 1{\sqrt x}\right)\sqrt{\sqrt x + \frac 1{\sqrt x}}} dx & \small \color{#3D99F6} \text{Let }u^2 = \sqrt x + \frac 1{\sqrt x} \\ & = \int \frac {4u}{\left(1+u^2\right)u} du & \small \color{#3D99F6} \implies 2u \ du = \left(\frac 1{2\sqrt x} - \frac 1{2x\sqrt x}\right)dx \\ & = \int \frac 1{1+u^2}du \\ & = 4\tan^{-1} u + C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \end{aligned}$

Therefore,

$g(x) = u = \sqrt{\sqrt x + \dfrac 1{\sqrt x}} \implies g^2(1) = \sqrt 1 + \frac 1{\sqrt 1} \implies \left \lfloor g^2(1) \right \rfloor = \boxed{2}$