Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 22

Evaluate $\displaystyle \int^9_4 \frac{\ln y}{\sqrt{y}} dy$ by using Integration by parts.
If we let $u = \ln y$ and $\displaystyle dv = \frac{dy}{\sqrt{y}}$,then
$\displaystyle du = \frac{1}{y} dy \text{ and } v = \int \frac{dy}{\sqrt{y}} = \frac{y^{-\frac{1}{2}+1} }{-\frac{1}{2}+1} = 2\sqrt{y}$

So,

$\begin{equation} \begin{aligned} \int^9_4 \frac{\ln y}{\sqrt{y}} dy = uv - \int vdu &= 2 \sqrt{y} \ln y - \int \frac{2\sqrt{y}dy}{y}\\ \\ &= 2 \sqrt{y} \ln y - \int 2 y^{\left( \frac{1}{2}-1\right)} dy\\ \\ &= 2 \sqrt{y} \ln y - 2 \int y^{-\frac{1}{2}} dy\\ \\ &= 2 \sqrt{y} \ln y - 2 (2 \sqrt{y})\\ \\ &= \left[ 2 \sqrt{y} (\ln y -2) \right]^9_4\\ \\ &= \left[ 2\sqrt{9} (\ln(9)-2) \right] - \left[ 2\sqrt{4}(\ln 4-2) \right]\\ \\ &= 6 \ln 9 - 12 - 4 \ln 4 + 8\\ \\ &= 6 \ln 9 - 4 \ln 4 - 4\\ \\ &= \ln 9^6 - \ln 4^4 - 4\\ \\ &= \ln \left( \frac{9^6}{4^4} \right) - 4\\ \\ &= \ln \left( \frac{531441}{256} \right) - 4 \end{aligned} \end{equation}$