Evaluating a simple looking integral

Question

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

Answer 1

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 $x$.

Answer 2

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 $$ \begin{aligned} I & =\frac{8}{\sqrt{3}} \int \frac{\tan ^2 \theta}{\sec ^2 \theta} d \theta \\ & =\frac{8}{\sqrt{3}}\int\frac{1-\cos 2 \theta}{2} d \theta \\ & =\frac{4}{\sqrt{3}}\left(\theta-\frac{\sin 2 \theta}{2}\right)+C \\ & =\frac{4}{\sqrt{3}}\arctan \sqrt{\frac{3 x+1}{3-3 x}}+\sqrt{\frac{3 x+1}{1-x}}(x-1)+ C \end{aligned} $$