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!
 A: 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.
$$\dfrac{2t^2}{(t^2-1)^2} = \dfrac12 \left(\dfrac1{(t-1)^2} + \dfrac1{(t-1)} + \dfrac1{(t+1)^2} - \dfrac1{(t+1)}\right)$$
A: 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*}
A: $$\int\sqrt{\frac{x}{x+1}}\,dx$$
Let $u^2=\frac{x}{x+1}=1-\frac{1}{x+1}$. Then $2u\, du=\frac{1}{(x+1)^2}dx$. So you have
$$\int2u^2(x+1)^2\,du=\int\frac{2u^2}{(1-u^2)^2}\,du$$
And it should be easy from there, as a rational function. Alternatively let $u=\sin t$ and you have
$$\int\frac{2\sin^2t}{\cos^{3}t}\,dt=\int2\tan^2t\sec t\,dt$$ if a trigonometric integral is preferable. Integration by parts on this last integral yields $$2\sec t\tan t-2\int\sec^3t\,dt$$ and there is a reduction formula for integrals of powers of $\sec t$.
A: 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.
A: Allow me to make an alternate suggestion
$$x=\tan^2t,dx=2\tan t\sec^2t$$
This yields
$$\int2\tan t\sec^2t\sqrt\frac{\tan^2t}{\tan^2t+1}dt=\int2\tan t\sec^2t\times\frac{\tan t}{\sec t}dt=$$
$$\int2\tan^2t\sec tdt=2\int\sec^3tdt-2\int\sec tdt$$
I assume you've seen these integrals before.  I believe the easiest way to handle the first half if I recall is to rewrite it as
$$\sec^3t=\frac1{\cos^3t}=\frac{\cos t}{\cos^4t}=\frac{\cos t}{(1-\sin^2t)^2}$$
