Integral Of $\int\sqrt{\frac{x}{x+1}}dx$ I want to solve this integral 
$$\int\sqrt{\frac{x}{x+1}}dx$$
And think about:
1) $t=\frac{x}{x+1}$

2) $dt = (\frac{1}{x+1} - \frac{x}{(x+1)^2})dx$ 
Now I need your advice! Thanks!

put
$$\sqrt{\frac{x}{x+1}}=t,\quad x=\frac{t^2}{1-t^2},\quad dx=\frac{2t}{(1-t^2)^2}\,\,dt$$
So
\begin{align*}
\int\sqrt{\frac{x}{x+1}}\,\,dx &=\int t\cdot \frac{2t}{(1-t^2)^2}\,\,dt=2\int  \frac{ t^2}{(1-t^2)^2}\,\,dt\\
&=2\int  \frac{ t^2+1-1}{(1-t)^2(1+t)^2}\,\,dt\\
&=2\int  \frac{ (t-1)(t+1)}{(1-t)^2(1+t)^2}\,\,dt-2\int  \frac{1}{(1-t)^2(1+t)^2}\,\,dt\\
&=-2\int  \frac{1}{ 1-t }\,\,dt-2\int  \frac{1}{(1-t)^2(1+t)^2}\,\,dt\\
&=-2\ln|1-t|-2\int  \frac{A}{1-t}+\frac{B}{(1-t)^2}+\frac{C}{1+t}+\frac{D}{(1+t)^2}\,\,dt\\
&=-2\ln|1-t|-2A\ln|1-t|-\frac{B}{ 1-t} -C\ln|1+t|+\frac{D}{ 1+t } 
\end{align*}
