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

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


$$\int \frac{1}{\sqrt{2x}-\sqrt{x+4}}\ dx=\int \frac{(\sqrt{2x}+\sqrt{x+4})}{(\sqrt{2x}-\sqrt{x+4})(\sqrt{2x}+\sqrt{x+4})}\ dx=$$ $$=\int \frac{\sqrt{2x}+\sqrt{x+4}}{x-4}\ dx$$ $$=\int \frac{\sqrt{2x}\ dx}{x-4} + \int \frac{\sqrt{x+4}\ dx}{x-4}$$ $$=\int \frac{\sqrt{2}\ xd(\sqrt{x})}{x-4} + \int \frac{2(x+4)d(\sqrt{x+4})}{x-4}$$ $$=\int \frac{(2\sqrt{2}(x-4)+8\sqrt2)d(\sqrt{x})}{x-4} + \int \frac{(2(x-4)+16)d(\sqrt{x+4})}{x-4}$$ $$=2\sqrt{2} \int d(\sqrt{x})+8\sqrt2\int \frac{d(\sqrt{x})}{(\sqrt{x})^2-2^2} + 2\int d(\sqrt{x+4})+16\int \frac{d(\sqrt{x+4})}{(\sqrt{x+4})^2-(2\sqrt2)^2}$$ $$=2\sqrt2\sqrt{x}+8\sqrt{2}\frac{1}{2\cdot 2}\ln\left|\frac{\sqrt x-2}{\sqrt{x}+2}\right|+2\sqrt{x+4}+16\frac{1}{2\cdot 2\sqrt2}\ln\left|\frac{\sqrt{x+4}-2\sqrt2}{\sqrt{x+4}+2\sqrt2}\right|$$ $$=2\sqrt{2x}+2\sqrt{2}\ln\left|\frac{\sqrt x-2}{\sqrt{x}+2}\right|+2\sqrt{x+4}+2\sqrt2\ln\left|\frac{\sqrt{x+4}-2\sqrt2}{\sqrt{x+4}+2\sqrt2}\right|+C$$


$$I=\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, dx$$

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

Tags:

Integration

Calculus

Indefinite Integrals

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