<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] If you know about Möbius transformations, from the change of variables
$$
t = \frac{x}{x+1}, \quad \Longrightarrow \quad x = \frac{t}{1-t}
$$
because
$$
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}.
$$
If you don't know about Möbius transformations, then you can do this : 
$$
t = \frac x{x+1} = 1 - \frac 1{x+1} \quad \Longrightarrow \quad 1-t = \frac 1{x+1} \quad \Longrightarrow \quad x = \frac 1{1-t} -1 = \frac t{1-t}.
$$This means that 
$$
\left( \frac 1{x+1} - \frac x{(1+x)^2} \right) = \frac{x+1-x}{(x+1)^2} = \frac 1{(x+1)^2} = \frac{1}{\left( \frac t{1-t} + 1 \right)^2} = (1-t)^2. 
$$
So now you can try to solve 
$$
\int \frac{\sqrt t}{(1-t)^2} \, dt.
$$
Letting $u = \sqrt t$, $du = \frac 1{2 \sqrt t} dt$ which implies $2u du = dt$, hence now we have 
$$
\int \frac{2u^2}{(1-u^2)^2} \, du
$$
to solve. Use partial fractions.
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