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?
 A: Using by parts actually gives you:
$$2\sqrt{(x)}\ln(x)-2\int\frac{1}{\sqrt{(x)}}dx$$
It's then fairly straightforward. 
A: Let $u(x)=\ln(x),v'(x)=\frac{1}{\sqrt x}$.
$$\int_a^9 \frac{\ln(x)}{\sqrt x}dx=\int_a^9 u(x)v'(x)dx=[u(x)v(x)]_a^9-\int_a^9u'(x)v(x)dx$$
$u'(x)=\frac{1}{x},v(x)=2\sqrt x$.
$$\begin{align}\int_a^9 \frac{\ln(x)}{\sqrt x}dx&=[\ln(x)2\sqrt x]_a^9-\int_a^9\frac{2\sqrt x}{x}dx\\&=\ln(9)2\sqrt9-\ln(a)2\sqrt a -4\int_a^9\frac{1}{2\sqrt x}dx\\&=\ln(9)2\sqrt9-\ln(a)2\sqrt a -4\sqrt 9+4\sqrt a \end{align} $$
When you make tend $a$ to $0$, you have : $$\int_0^9 \frac{\ln(x)}{\sqrt x}dx=\ln(9)2\sqrt9-0 -4\sqrt 9+0=2\sqrt 9(\ln9-2)$$
A: $\frac{ln(x)}{\sqrt{x}}=2\frac{ln(\sqrt{x})}{\sqrt{x}}$ now put $\sqrt{x}=u$ so integral becomes $4.ln(u).du$ which can easily be integrated.
A: 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}
A: : Let $u=\ln x \,\,\And\,\,  {{e}^{u}}=x\Leftrightarrow du=\frac{dx}{x}\Leftrightarrow {{e}^{u}}du=dx$ ,then
$\int_{0}^{9}{\frac{\ln x}{\sqrt{x}}dx}=\int_{-\infty }^{\ln 9}{\frac{u{{e}^{u}}du}{\sqrt{{{e}^{u}}}}}=\int_{-\infty }^{\ln 9}{u{{e}^{u\left( 1-1/2 \right)}}du}=\int_{-\infty }^{\ln 9}{u\sqrt{{{e}^{u}}}du}$ 
Let $U=u\,\,\And \,\,dV=\sqrt{{{e}^{u}}}du\Leftrightarrow dU=du\,\,\And \,\,V=2\sqrt{{{e}^{u}}}$then 
$\int_{-\infty }^{\ln 9}{u\sqrt{{{e}^{u}}}}du=\left[ 2u\sqrt{{{e}^{u}}} \right]_{-\infty }^{\ln 9}-2\int_{-\infty }^{\ln 9}{\sqrt{{{e}^{u}}}du=\left[ 2u\sqrt{{{e}^{u}}}-4\sqrt{{{e}^{u}}} \right]_{-\infty }^{\ln 9}=\left[ \sqrt{{{e}^{u}}}\left( 2u-4 \right) \right]_{-\infty }^{\ln 9}}$
$\therefore \int_{0}^{9}{\frac{\ln x}{\sqrt{x}}dx}=\underset{u\to -\infty }{\mathop{\lim }}\,\left( \sqrt{{{e}^{\ln 9}}}\left( 2\ln 9-4 \right)-\sqrt{{{e}^{u}}}\left( 2u-4 \right) \right)=\left( 3\left( 4\ln 3-4 \right) \right)=12\left( \ln 3-1 \right)$ 
