<s>[INST] 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.
 [/INST] 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 $y$.
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