How many ways to deal with the integral $\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}$? I tackle the integral by rationalization on the integrand first.
$$
\frac{1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2 x}
$$
Then splitting into two simpler integrals yields  $$
\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{1}{2}\left [\underbrace{\int\frac{\sqrt{1+x}}{x}}_{J} d x+\underbrace{\int\frac{\sqrt{1-x}}{x} d x}_{K}\right]
$$
To deal with $J$, we use rationalization instead of substitution. $$
\begin{aligned}
J &=\int \frac{\sqrt{1+x}}{x} d x \\
&=\int \frac{1+x}{x \sqrt{1+x}} d x \\
&=2 \int\left(\frac{1}{x}+1\right) d(\sqrt{1+x}) \\
&=2 \int \frac{d(\sqrt{1+x})}{x}+2 \sqrt{1+x} \\
&=2 \int \frac{d(\sqrt{1+x})}{(\sqrt{1+x})^{2}-1}+2 \sqrt{1+x} \\
&=\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1} \right| +2 \sqrt{1+x}+C_{1}
\end{aligned}
$$
$\text {Replacing } x \text { by } -x \text { yields }$
$$
\begin{array}{l} \\
\displaystyle K=\int \frac{\sqrt{1-x}}{-x}(-d x)=\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|+2 \sqrt{1-x}+C_{2}
\end{array}
$$
Now we can conclude that $$
I=\sqrt{1+x}+\sqrt{1-x}+\frac{1}{2}\left(\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|\right)+C
$$
My question is whether there are  any simpler methods such as integration by parts , trigonometric substitution, etc…
Please help if you have. Thank you for your attention.
 A: Seek $f$ so$$\frac{\sqrt{1+x}+\sqrt{1-x}}{2x}=f(1+x)-f(1-x),$$e.g.$$f(y):=\frac{\sqrt{y}}{2(y-1)}.$$You want to evaluate$$\int(f(1+x)-f(1-x))dx=F(1+x)+F(1-x)+C$$with $F^\prime(y)=F(y)$.
Just about any concise solution technique will exploit the above facts. Your approach is equivalent to next taking $y=z^2$, so$$F(y)=\int\frac12\left(2+\frac{1}{z-1}-\frac{1}{z+1}\right)dz=\sqrt{y}+\frac12\ln\left|\frac{\sqrt{y}-1}{\sqrt{y}+1}\right|+C.$$I doubt there's anything much simpler than this, but what's preferable is up to taste. In terms of trigonometric substitutions, you may like $y=\cos^2u$ so$$fdy=\frac12(\csc u-\sin u)du$$or $y=\sec^2u$ so$$fdy=\frac12(\cot u+\tan u)du,$$depending on the range of $y$.
A: Let $x=\cos 2 \theta$, for $0\leq 2\theta < \pi$, then $d x=-2 \sin 2 \theta d \theta$ and
$$
\begin{aligned}
I &=\int \frac{-2 \sin 2 \theta d \theta}{\sqrt{1+\cos 2 \theta-\sqrt{1-\cos 2 \theta}}} \\
&=-2 \int \frac{\sin 2 \theta d \theta}{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}} \\
&=-\sqrt{2} \int \frac{\sin 2 \theta}{\cos \theta-\sin \theta} \cdot d \theta \\
&=-\sqrt{2} \int \frac{1-(\cos \theta-\sin \theta)^{2}}{\cos \theta-\sin \theta} d \theta \\
&=\sqrt{2}\left[\underbrace{\int\frac{d \theta}{\sin \theta-\cos \theta}}_{J}+\underbrace{\int(\cos \theta-\sin \theta) d \theta}_{K}\right]
\end{aligned}
$$
$$
\begin{array}{l}
\displaystyle J=\frac{1}{\sqrt{2}} \ln \left|\frac{\sin \theta-\cos \theta}{\sqrt{2}+\cos \theta+\sin \theta}\right|+C_{1} \\
\displaystyle I=\ln \left|\frac{\sin \theta-\cos \theta}{\sqrt{2}+\cos \theta+\sin \theta}\right|+\sqrt{2}(\sin \theta+\cos \theta)+C
\end{array}
$$
Since  $0\leq \theta < \dfrac{\pi}{2}$, putting $ \displaystyle \sin \theta=\sqrt{\frac{1-\cos 2 \theta}{2}}=\sqrt{\frac{1-x}{2}}$ $\textrm{ and }$
$\displaystyle \cos \theta=\sqrt{\frac{1+x}{2}}$ yields
$$
I=\sqrt{1-x}+\sqrt{1+x}+\ln \left|\frac{\sqrt{1-x}-\sqrt{1+x}}{2+\sqrt{1+x}+\sqrt{1-x}}\right|+C
$$
A: Here is another approach, $$\int\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2x}dx=\int\dfrac{\sqrt{2+2\sqrt{1-x^2}}}{2x}dx=\frac1{\sqrt2}\int\dfrac{\sqrt{1+\sqrt{1-x^2}}}xdx$$
Let, $x=\sin2\theta$,
$$\int\dfrac{\cos\theta}{\sin2\theta}\times2\cos2\theta \;d\theta=\int\dfrac{\cos2\theta}{\sin\theta}d\theta=\int\csc\theta-2\sin\theta \;d\theta$$$$=\ln|\csc\theta-\cot\theta|+2\cos\theta+C$$
Now we need to denote the result in term of $x$,$\cos2\theta=\sqrt{1-x^2}=2\cos^2\theta-1\;\Rightarrow\;2\cos\theta=\sqrt{2+2\sqrt{1-x^2}}$
$\csc\theta-\cot\theta=\dfrac{1-\cos\theta}{\sin\theta}=\dfrac{2\cos\theta-\cos2\theta-1}{\sin2\theta}=\dfrac{\sqrt{2+2\sqrt{1-x^2}}-\sqrt{1-x^2}-1}{x}$
Hence, $$\int\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2x}dx=\sqrt{2+2\sqrt{1-x^2}}+\ln\left|\dfrac{\sqrt{2+2\sqrt{1-x^2}}-\sqrt{1-x^2}-1}{x}\right|+C$$
A: The answer is no.
There are only two different representations of the result at all.
$ 1/2 (2 \sqrt{1-x}+2 \sqrt{x+1}+ln(\sqrt{1-x}-1)-ln(\sqrt{1-x}]+1)+ln(\sqrt{x+1}-1)-ln(\sqrt{x+1}+1))+constant$
or alternatively:
$ \sqrt{1-x}+\sqrt{x+1}+1/2 ln(((\sqrt{1-x}-1) (\sqrt{x+1}-1))/((\sqrt{1-x}+1) (\sqrt{x+1}+1)))+constant$
This can be rewritten as:
$\sqrt{1-x}+\sqrt{x+1}-tanh^{-1}(\sqrt{1-x})-tanh^{-1}(\sqrt{x+1})+constant$
All valid on the domains where the given function is valid and the individual terms are valid for example real.
There is no other way to integrate the given function. Like the example by Etemon and J.G. shows it is really important to obey for each of the substitutions the domains on/in which they are valid. There are several branches that do not lead to a proper integration.
The integration path is:
(1) $u=\sqrt{x+1}$
(2) $s=u+1$
(3) $p=u-1$
(4) $w=\sqrt{1-x}$
(5) $v=w+1$
(6) $z=w-1$
And the long divisions in between. So written this in length is much longer if done correctly than Your path. Look for example in the step-by-step solution of Wolfram Alpha or Maplesoft Online. Which are both free or the given one by Mikasa. These are standardized digital step-by-step solution conform with suggestions of the NIST. NIST digests are targeted for efficiency and accurateness. These are usually unique integration solution in step-by-step format.
There is only the identity between $ln$ and $arctanh$ that can be used. Six steps and backsubstitution is really long and tedious.
