Answer
Diverges
Work Step by Step
$\int\limits_{1}^{\infty}\frac{1+Sin^2x}{\sqrt x}\, dx$
By Squeeze theorem, $0 \leq Sin^2x \leq 1$,
Therefore, minimum value of the integral is given by : $\int\limits_{1}^{\infty}\frac{1+0}{\sqrt x}\, dx$ = $\int\limits_{1}^{\infty}\frac{1}{\sqrt x}\, dx$
Multiplying top and bottom by 2:
$2\int\limits_{1}^{\infty}\frac{1}{2\sqrt x}\, dx$ = $2 \sqrt x$ $\Big| _{1}^{\infty}$ = $2(\sqrt{\infty} - \sqrt1)=\infty$ .
