Evaluate $\int_0^9 \frac{\ln(x)}{\sqrt x}$ I need to solve this defined integral:

$$\int_0^9 \frac{\ln(x)}{\sqrt{x}}dx$$

I tried to solve by parts, so:
$$f(x)g(x) = \int f'(x)g(x)dx$$
where:
$$f(x) = \ln(x)$$
$$f'(x) = \frac{1}{x}$$
$$g'(x) = \frac{1}{\sqrt{x}}$$
$$g(x) = 2\sqrt{x}$$
And I get:
$$\ln(x)2\sqrt x - \int\frac{1}{x} \cdot \frac{1}{\sqrt x}$$
How should I proceed?

Consider
\begin{equation}
I(a) = \int\limits_{0}^{b} x^{a} \mathrm{d} x = \frac{x^{a+1}}{a+1}\bigg|_{0}^{b} = \frac{b^{a+1}}{a+1}
\end{equation}
Then
\begin{equation}
\int\limits_{0}^{b} x^{a} \mathrm{ln}(x) \mathrm{d} x = \frac{\partial I(a)}{\partial a} = 
\frac{b^{a+1}}{(a+1)^{2}}[(a+1)\mathrm{ln}(b) - 1]
\end{equation}
and
\begin{align}
\int\limits_{0}^{b} \frac{\mathrm{ln}(x)}{\sqrt{x}} \mathrm{d} x & = 
\lim_{a \to -\frac{-1}{2}} \int\limits_{0}^{b} x^{a} \mathrm{ln}(x) \mathrm{d} x \\
& = \lim_{a \to -\frac{-1}{2}} \frac{b^{a+1}}{(a+1)^{2}}[(a+1)\mathrm{ln}(b) - 1] \\
& = 4\sqrt{b}[\mathrm{ln}(\sqrt{b})-1]
\end{align}
For $b=9$ we have
\begin{equation}
\int\limits_{0}^{9} \frac{\mathrm{ln}(x)}{\sqrt{x}} \mathrm{d} x
= 4\sqrt{9}[\mathrm{ln}(\sqrt{9})-1]
= 12(\mathrm{ln}3-1)
\end{equation}
