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.

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 $y$.

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 $$ I=\sqrt{1-x}+\sqrt{1+x}+\ln \left|\frac{\sqrt{1-x}-\sqrt{1+x}}{2+\sqrt{1+x}+\sqrt{1-x}}\right|+C $$