How to solve $intfrac{1}{sqrt {2x} - sqrt {x+4}} , mathrm{dx} $?

$$I=intfrac{1}{sqrt {2x} - sqrt {x+4}} , dx$$

Get rid of the denominator $$u=sqrt {2x} - sqrt {x+4}implies x=3 u^2+2 sqrt{2}u sqrt{u^2+4 }+4implies frac{dx}{du}=frac{sqrt{2} left(4 u^2+8 right)}{sqrt{u^2+4 }}+6 u$$ $$I=int6 du+intfrac{4 sqrt{2} u}{sqrt{u^2+4}}du+intfrac{8 sqrt{2}}{usqrt{u^2+4} }du$$ $$I=6u+4 sqrt{2} sqrt{u^2+4}-4 sqrt{2} log left(frac{sqrt{u^2+4}+2}{u}right)$$

$$int frac{1}{sqrt{2x}-sqrt{x+4}} dx=int frac{(sqrt{2x}+sqrt{x+4})}{(sqrt{2x}-sqrt{x+4})(sqrt{2x}+sqrt{x+4})} dx=$$ $$=int frac{sqrt{2x}+sqrt{x+4}}{x-4} dx$$ $$=int frac{sqrt{2x} dx}{x-4} + int frac{sqrt{x+4} dx}{x-4}$$ $$=int frac{sqrt{2} xd(sqrt{x})}{x-4} + int frac{2(x+4)d(sqrt{x+4})}{x-4}$$ $$=int frac{(2sqrt{2}(x-4)+8sqrt2)d(sqrt{x})}{x-4} + int frac{(2(x-4)+16)d(sqrt{x+4})}{x-4}$$ $$=2sqrt{2} int d(sqrt{x})+8sqrt2int frac{d(sqrt{x})}{(sqrt{x})^2-2^2} + 2int d(sqrt{x+4})+16int frac{d(sqrt{x+4})}{(sqrt{x+4})^2-(2sqrt2)^2}$$ $$=2sqrt2sqrt{x}+8sqrt{2}frac{1}{2cdot 2}lnleft|frac{sqrt x-2}{sqrt{x}+2}right|+2sqrt{x+4}+16frac{1}{2cdot 2sqrt2}lnleft|frac{sqrt{x+4}-2sqrt2}{sqrt{x+4}+2sqrt2}right|$$ $$=2sqrt{2x}+2sqrt{2}lnleft|frac{sqrt x-2}{sqrt{x}+2}right|+2sqrt{x+4}+2sqrt2lnleft|frac{sqrt{x+4}-2sqrt2}{sqrt{x+4}+2sqrt2}right|+C$$

Multiplying by the conjugate and applying a couple of substitutions does work. begin{align*} intfrac{1}{sqrt {2x} - sqrt {x+4}} , mathrm{d}x &=int underbrace{frac{sqrt{2x}}{x-4}}_{sqrt{x} to u} + underbrace{frac{sqrt{x+4}}{x-4}}_{sqrt{x+4} to t} ; mathrm{d}x\ &=2sqrt{2} int frac{u^2}{u^2-4} ; mathrm{d}u+ 2int frac{t^2}{t^2-8} ; mathrm{d}t \ &=2sqrt{2} int frac{u^2-4+4}{u^2-4} ; mathrm{d}u+ 2int frac{t^2-8+8}{t^2-8} ; mathrm{d}t \ &=2sqrt{2}u +2sqrt{2} ln{bigg |frac{u-2}{u+2}bigg |} + 2t +2sqrt{2}ln{bigg |frac{t-2sqrt{2}}{t+2sqrt{2}}bigg |}+mathrm{C} \ &=2sqrt{2x} +2sqrt{2} ln{bigg |frac{sqrt{x}-2}{sqrt{x}+2}bigg |} + 2sqrt{x+4} +2sqrt{2}ln{bigg |frac{sqrt{x+4}-2sqrt{2}}{sqrt{x+4}+2sqrt{2}}bigg |}+mathrm{C} \ end{align*}