Rewrite $\frac{1}{\sqrt{x}}$ as $x^{-\frac{1}{2}}$. Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{-\frac{1}{2}}\mathrm{d}x$ with $\frac{x^{\frac{1}{2}}}{\frac{1}{2}}$. Simplify and convert from exponential to radical form. Multiply $8$ times $2\sqrt{x}$.
$$16\sqrt{x}-3\int \sqrt{x}\mathrm{d}x$$
Rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$. Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{\frac{1}{2}}\mathrm{d}x$ with $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}$. Simplify. Multiply $-3$ times $\frac{2x^{\frac{3}{2}}}{3}$.
$$16\sqrt{x}-2x^{\frac{3}{2}}$$
The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.
