Calculate simple integral undefined I'm not being able to calculate $ \large{\int{\sqrt{\frac{x}{a-x}}} dx} $ , someone could help me? I tryed to use integration by parts, but i achieved $0 = 0$.  
Thanks in advance.

Integrals of roots (of any order) of a linear function can be integrated by substitution, in this case of 
$$y^2=\frac{x}{a-x}$$
solving for $x$ we get $$x=\frac{ay^2}{y^2+1}$$ so
$$dx=\frac{2ay}{(y^2+1)^2}dy$$
substituting this gives,
$$\int\sqrt{\frac{x}{a-x}}dx=\int \frac{2ay^2}{(y^2+1)^2}dy$$
which is a rational function and can be integrated by partial fractions.
In this case
$$\int \frac{y^2}{(y^2+1)^2}dy=\int \frac{1}{y^2+1}dy-\int \frac{1}{(y^2+1)^2}dy$$
$$\int \frac{1}{y^2+1}dy=\arctan(y)$$ is easy but the other integral is harder and can be done in several ways including trig substitution, but can also be done directly by parts
\begin{equation*}
\begin{split}
\int \frac{1}{y^2+1}dy&=\frac{y}{y^2+1}-\int \frac{2y^2}{(y^2+1)^2}dy\\
&=\frac{y}{y^2+1}-2\int \frac{1}{y^2+1}dy  +2\int \frac{1}{(y^2+1)^2}dy\\
\end{split}
\end{equation*}
So
\begin{equation*}
\begin{split}
2\int \frac{1}{(y^2+1)^2}dy&=3\int \frac{1}{y^2+1}dy  -\frac{y}{y^2+1}\\
&=3\int \arctan(y)  -\frac{y}{y^2+1}\\
\end{split}
\end{equation*}
And we have 
$$\int \frac{y^2}{(y^2+1)^2}dy= -\frac{1}{2}\int \arctan(y)  +\frac{1}{2}\frac{y}{y^2+1}$$
Then assuming I have not made any mistakes, (which is unlikely) you should be able to show it is equal to the Wolfram answer.
