Make a substitution to express the integrand as a rational function and then evaluate the integral.
integral from 16 to 9 of (square root x)/ (x-4) dx (Let u= square root of x.)
Thanks
$\displaystyle \int_9^{16} \frac{\sqrt{x}}{x-4}\,dx$
If $\displaystyle u = \sqrt{x}$ then $\displaystyle du = \frac{dx}{2\sqrt{x}}$ or $\displaystyle 2 u du = dx$
So (remember - change your limits of integration)
$\displaystyle \int_9^{16} \frac{\sqrt{x}}{x-4}\,dx = \int_3^4 \frac{2u^2}{u^2-4}\, du$. Can you take it from here?