<s>[INST] 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!
 [/INST] Let $$\sqrt{\dfrac{x}{x+1}} = t \implies \dfrac{x}{x+1} = t^2 \implies x+1 = \dfrac1{1-t^2} \implies dx = \dfrac{2tdt}{(t^2-1)^2}$$
Hence,
$$\int \sqrt{\dfrac{x}{x+1}} dx = \int \dfrac{2t^2}{(t^2-1)^2}dt$$
I trust you can take it from here using partial fractions.
$$\dfrac{2t^2}{(t^2-1)^2} = \dfrac12 \left(\dfrac1{(t-1)^2} + \dfrac1{(t-1)} + \dfrac1{(t+1)^2} - \dfrac1{(t+1)}\right)$$
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