How to integrate $\int\frac{dx}{\sqrt{x+1}+\sqrt{x-1}}$? Firstly, I tried to multiply the denominator and numerator by $\sqrt{x+1}$ but to no avail. Then I tried to take $\sqrt{x+1} = u$; now $\sqrt{x-1}=u-2$ but I couldn't find any success here either.
How can I integrate this expression?
 A: Hint: Rationalise the denominator; that is, multiply the integrand by $\dfrac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}$.
A: I shall continue one step further from @A. Goodier's answer:
Multiplying the numerator and denominator by the conjugate, we have: 
\begin{align}
   \int \frac{\mathrm{d}x}{\sqrt{x+1}+\sqrt{x-1}}&=\int\frac{\sqrt{x+1}-\sqrt{x-1}}{(x+1)-(x-1)}\mathrm{d}x\\
&=\frac12\int\sqrt{x+1}-\sqrt{x-1}\ \ \mathrm{d}x
\end{align}
This is a simple integral and can easily be solved using the standard formulae.
A: Setting $\small\left\lbrace\begin{aligned}x&=\cosh{\left(2y\right)}\\ \mathrm{d}x&=2\sinh{\left(2y\right)}\,\mathrm{d}y\end{aligned}\right. $, we get : \begin{aligned}\int{\frac{\mathrm{d}x}{\sqrt{1+x}+\sqrt{x-1}}}&=\frac{1}{\sqrt{2}}\int{\frac{2\sinh{\left(2y\right)}}{\sinh{y}+\cosh{y}}\,\mathrm{d}y}\\ &=\frac{1}{\sqrt{2}}\int{\mathrm{e}^{-y}\left(\mathrm{e}^{2y}-\mathrm{e}^{-2y}\right)\mathrm{d}y}\\ &=\frac{1}{\sqrt{2}}\int{\mathrm{e}^{y}\,\mathrm{d}y}-\frac{1}{\sqrt{2}}\int{\mathrm{e}^{-3y}\,\mathrm{d}y}\\ &=\frac{\mathrm{e}^{y}}{\sqrt{2}}+\frac{\mathrm{e}^{-3y}}{3\sqrt{2}}+C\\ &=\frac{\mathrm{e}^{\frac{\cosh^{-1}{\left(x\right)}}{2}}}{\sqrt{2}}+\frac{\mathrm{e}^{-\frac{3\cosh^{-1}{\left(x\right)}}{2}}}{3\sqrt{2}}+C\\ &=\frac{1}{\sqrt{2}}\sqrt{x+\sqrt{x^{2}-1}}+\frac{1}{3\sqrt{2}\left(x+\sqrt{x^{2}-1}\right)\sqrt{x+\sqrt{x^{2}-1}}}+C\\ &=\frac{\sqrt{x+1}+\sqrt{x-1}}{2}+\frac{1}{3\left(x+\sqrt{x^{2}-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}+C\end{aligned}
