Derivation of$~\int\sqrt{\frac{x-1}{x+1}}\,dx=4\int\frac{t^2}{\left(1-t^2\right)^2}\,dt~~\text{where}~~~t:=\sqrt{\frac{x-1}{x+1}}~$

Question

$$\underbrace{\int_{}^{}\sqrt{\frac{x-1}{x+1}}\,dx=4\int\frac{t^2}{\left(1-t^2\right)^2}\,dt}_\text{I want to derive RHS from LHS}~~\text{where}~~~t:=\sqrt{\frac{x-1}{x+1}}\tag{1}$$

$$x\neq-1\tag{2}$$

$$A:=\int_{}^{}\sqrt{\frac{x-1}{x+1}}\,dx\tag{3}$$

$$t=\sqrt{\frac{x-1}{x+1}}\tag{4}$$

$$=\frac{\sqrt{x-1}}{\sqrt{x+1}}\tag{5}$$

$$t^2=\frac{x-1}{x+1}\tag{6}$$

$$\underbrace{\left(x+1\right)t^2=\left(x-1\right)}_{\text{I will}~\frac{d}{dx}~\text{each formula}}~\tag{7}$$

$$\frac{d}{dx}\left(\left(x+1\right)t^2\right)=\frac{d}{dx}\left(\left(x-1\right)\right)\tag{8}$$

$$t^2+\left(x+1\right)\frac{d}{dx}\left(t^2\right)=1\tag{9}$$

$$t^2+\left(x+1\right)2t\frac{dt}{dx}=1\tag{10}$$

$$2t\left(x+1\right)\frac{dt}{dx}=1-t^2\tag{11}$$

$$\frac{dt}{dx}=\frac{1-t^2}{2t\left(x+1\right)}\tag{12}$$

$$A=\int_{}^{}\sqrt{\frac{x-1}{x+1}}\,dx\tag{13}$$

$$\frac{dx}{dt}=\frac{2t\left(x+1\right)}{\left(1-t^2\right)}\tag{14}$$

$$dx=\frac{2t\left(x+1\right)}{\left(1-t^2\right)}dt\tag{15}$$

$$\therefore~~A=\int_{}^{}\frac{\sqrt{x-1}}{\sqrt{x+1}}\cdot\frac{2t\left(x+1\right)}{\left(1-t^2\right)}\,dt\tag{16}$$

$$=2\int_{}^{}\frac{t\sqrt{x-1}}{\left(1-t^2\right)}\frac{\left(x+1\right)}{\sqrt{x+1}}\,dt\tag{17}$$

$$=2\int_{}^{}\frac{t}{1-t^2}\cdot\frac{\sqrt{x-1}}{1}\cdot\frac{\left(x+1\right)}{\sqrt{x+1}}\cdot\frac{\sqrt{x+1}}{\sqrt{x+1}}\,dt\tag{18}$$

$$=2\int_{}^{}\frac{t}{1-t^2}\cdot\frac{\sqrt{x-1}}{1}\cdot\sqrt{x+1}\,dt\tag{19}$$

$$=2\int_{}^{}\frac{t}{1-t^2}\cdot\sqrt{x^2-1}\,dt\tag{20}$$

How should I handle$~\sqrt{x^2-1}~$?

ADD

After read this almost same problem, I've got the following.

$$x=\frac{\left(1+t^2\right)}{\left(1-t^2\right)}\tag{21}$$

$$x=\left(1+t^2\right)\left(1-t^2\right)^{-1}\tag{22}$$

$$\frac{dx}{dt}=\left(2t\right)\left(1-t^2\right)^{-1}+\left(1+t^2\right)\left(-1\right)\left(1-t^2\right)^{-2}\left(-2t\right)\frac{dt}{dx}\tag{23}$$

Stuked again from here.

ADD2

After read commentary of Gary, I got the following WIP.

$$\sqrt{x^2-1}\tag{24}$$

$$=\sqrt{\left(\frac{1+t^2}{1-t^2}\right)^2-1}\tag{25}$$

$$\sqrt{\frac{\left(1+t^2\right)^2}{\left(1-t^2\right)^2}-1}\tag{26}$$

$$=\sqrt{\frac{\left(1+t^2\right)^2-\left(1-t^2\right)^2}{\left(1-t^2\right)^2}}\tag{27}$$

$$=\sqrt{\frac{\left(t^2+1\right)^{2}-\left(t^2-1\right)^{2}}{\left(t^2-1\right)^2}}\tag{28}$$

$$=\frac{1}{\sqrt{\left(t^2-1\right)^{2}}}\cdot\sqrt{\left(t^2+1\right)^2-\left(t^2-1\right)^2}\tag{29}$$

$$=\frac{1}{\left|t^2-1\right|}\cdot\sqrt{t^4+2t^2+1-\left(t^4-2t^2+1\right)}\tag{30}$$

$$=\frac{1}{\left|t^2-1\right|}\cdot\sqrt{t^4+2t^2+1-t^4+2t^2-1}\tag{31}$$

$$=\frac{1}{\left|t^2-1\right|}\sqrt{4t^2}\tag{32}$$

$$=\frac{\sqrt{4}\sqrt{t^2}}{\left|t^2-1\right|}\tag{33}$$

$$=\frac{2\left|t\right|}{\left|t^2-1\right|}=:B\tag{34}$$

$$t^2\leq1~~\leftarrow~~\because\text{proof by contradiction}\tag{35}$$

$$\therefore~~B=\frac{2\left|t\right|}{1-t^2}\tag{36}$$

The remaining problem is to prove why absolute signs can be removed above.

I think I solved it WIP

Answer 1

You can go from the RHS to the LHS...

\begin{align*} 4\int \frac{t^2}{(1-t^2)^2}\,dt = & 4\int \dfrac{\frac{x-1}{x+1}}{\left(1-\frac{x-1}{x+1}\right)^2}\cdot \dfrac{1}{\sqrt{\frac{x-1}{x+1}}(1+x)^2} dx \\ = & \int \sqrt{\frac{x-1}{x+1}}\,dx \end{align*}

Or, from the LHS to the RHS:

Since you can easily see that $\displaystyle x = \frac{1+t^2}{1-t^2}, \quad \frac{dx}{dt} = \frac{4t}{(1-t^2)^2}$, you have that

$$ \int \sqrt{\frac{x-1}{x+1}}\,dx = \int t \cdot \frac{4t}{(1-t^2)^2}\, dt = 4 \int \frac{t^2}{(1-t^2)^2}\, dt. $$

Answer 2

Notice that if $x<-1$ or $x\geq 1$ $$ t = \sqrt{\frac{x-1}{x+1}} \iff (x+1) \,t^2 = x-1 \\ \iff x = \frac{1+t^2}{1-t^2} $$ Therefore $\mathrm d x = \frac{4\,t}{(1-t^2)^2}\, \mathrm d t$ and so $$ \int\sqrt{\frac{x-1}{x+1}} \mathrm d x = \int \frac{4\,t^2}{(1-t^2)^2}\mathrm d t $$

Answer 3

$$t=\sqrt{\frac{x-1}{x+1}}\tag{1}$$

I will use method of proof of contradiction.

$$\text{Assumpted}~~~\frac{x-1}{x+1}\geq1\tag{2}$$

$$\left(x-1\right)\geq\left(x+1\right)~~\leftarrow~~\text{Error exists}\tag{3}$$

$$\therefore~~\frac{x-1}{x+1}\leq1\tag{4}$$

The remaining concern is whether$~t~$be imaginary number.

$$x=1~~\Leftrightarrow~~t=0\tag{5}$$

$$x>1~~\Leftrightarrow~~\frac{x-1}{x+1}>0~~\Leftrightarrow~~t>0\tag{6}$$

$$x<-1~~\Leftrightarrow~~\frac{x-1}{x+1}>0~~\Leftrightarrow~~t>0\tag{7}$$

$$-1<x<1~~\Leftrightarrow~~\left|x\right|<1\tag{8}$$

As$~\left|x\right|<1~$is held then$~\frac{x-1}{x+1}<0~$is satisfied.

WIP