the integral from 0 to 1 of lnx/(x^1/2)
thanks!
try integration by parts...
$\displaystyle u=ln(x)$ and $\displaystyle dv=\frac{1}{\sqrt{x}}dx$
$\displaystyle du=\frac{1}{x}dx$ and $\displaystyle v=2\sqrt{x}$
so we get
$\displaystyle \int\frac{ln(x)}{\sqrt{x}}=2\sqrt{x}ln(x)-\int 2 \sqrt{x}\frac{1}{x}dx$
$\displaystyle \int\frac{ln(x)}{\sqrt{x}}=2\sqrt{x}ln(x)-2\int x^{-1/2}dx$
You should be able to get it from here.
I'm pretty sure that luckeyduck is not ready to digest the following solution, but I'd like to show this attempt, since double integration tricks make (sometimes) these things easier:
$\displaystyle \begin{aligned}
\int_0^1 {\frac{{\ln x}}
{{\sqrt x }}\,dx} &= - \int_0^1 {\int_x^1 {\frac{{dy\,dx}}
{{y\sqrt x }}} } \hfill \\
&= - \int_0^1 {\underbrace {\int_0^y {\frac{{dx}}
{{\sqrt x }}} }_{2\sqrt y }\,\frac{{dy}}
{y}} \hfill \\
&= - 2\int_0^1 {\frac{1}
{{\sqrt y }}\,dy} \hfill \\
&= - 4.
\end{aligned}$