Evaluating a simple looking integral While solving an problem of Reator design i have encountered a integral,
$$ I =\int \left(\frac{3x+1}{1-x}\right)^{1/2} dx$$
$0<x<1$
My Approach: Rationalization didn't helped much, so i have substituted $$\frac{3x+1}{1-x}=t $$
and $$dt =\frac{3(1-x)+(3x+1)}{(1-x)^2}dx= \frac{4}{(1-x)^2}dx $$
also $x$ would become $x = \frac{t-1}{t+3}$,
the integral would become
$$I = 2\int \sqrt{t}\left(\frac{t+1}{(t+3)^2}\right) dt $$
$$I =2\int \left[\frac{t^{3/2}}{(t+3)^2} + \frac{t^{1/2}}{(t+3)^2}\right]dt $$
From here i can't form the expression into some standard integrable form,how can i proceed further.
Hints are appreciated
 A: First substitute $t=1-x$, which gives $$-\int (\frac{4-3t}{t})^{1/2}dt$$. Then substitute $t=\frac{4}{3}\sin^2u$ which gives $$-\frac{8}{3}\int \frac{2 \cos u}{(2/\sqrt 3)\sin u}\sin u \cos udu$$ You can easily integrate in terms of $u$ and then substitute to express the result first in terms of $t$ and then in terms of $x$.
A: We may deal with the integral by a single substitution
$$
\sqrt{3} \tan \theta=\left(\frac{3 x+1}{1-x}\right)^{\frac{1}{2}},
$$
then
$$
\begin{aligned}
6 \tan \theta \sec ^2 \theta d \theta & =\frac{4}{(1-x)^2} d x \\
& =4\left(\frac{3+3 \tan ^2 \theta}{4}\right)^2 d x \\
& =\frac{9}{4} \sec ^4 \theta d x
\end{aligned}
$$
and the integral can be transformed into
$$
\begin{aligned}
I & =\frac{8}{\sqrt{3}} \int \frac{\tan ^2 \theta}{\sec ^2 \theta} d \theta \\
& =\frac{8}{\sqrt{3}}\int\frac{1-\cos 2 \theta}{2} d \theta \\
& =\frac{4}{\sqrt{3}}\left(\theta-\frac{\sin 2 \theta}{2}\right)+C \\
& =\frac{4}{\sqrt{3}}\arctan \sqrt{\frac{3 x+1}{3-3 x}}+\sqrt{\frac{3 x+1}{1-x}}(x-1)+ C
\end{aligned}
$$
