Evaluate the following integrals:

Given:

$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x$

By applying $(a-b)^{2}=a^{2}-2 a b+b^{2}$

$\Rightarrow \int\left((\sqrt{x})^{2}+\left(\frac{1}{\sqrt{x}}\right)^{2}-2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right)\right) d x$

$\Rightarrow \int\left((\sqrt{x})^{2}+\left(\frac{1}{\sqrt{x}}\right)^{2}-2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right)\right) d x$

After computing,

$\Rightarrow \int\left(x+\frac{1}{x}-2\right) d x$

By Splitting, we get,

$\int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}$

$\int\left(\frac{1}{\mathrm{x}}\right) \mathrm{d} \mathrm{x}=\log \mathrm{x}$

$\int \mathrm{kdx}=\mathrm{kx}+\mathrm{c}$

We get,

$\Rightarrow \frac{x^{1+1}}{1+1}+\log x-2 x+c^{l}=1 / 2 x^{2}+\log x-2 x+c$