Question

$\int \frac{1}{x^{2/3}\sqrt{x^{2/3}-4}}dx$

Answer 1

$$\begin{gathered} \int \frac{dx}{x^{{\displaystyle \frac{2}{3}}}\sqrt{x^{{\displaystyle \frac{2}{3}}}-2^2}} \\ =\int \frac{dx}{x^{{\displaystyle \frac{2}{3}}}\sqrt{{\left(x^{{\displaystyle \frac{1}{3}}}\right)}^2-2^2}} \\ \mathrm{Let}{x}^{\frac{1}{3}}=t \\ \Rightarrow \frac{1}{3}{x}^{\frac{-2}{3}}dx=dt \\ \Rightarrow \frac{1}{3x^{{\displaystyle \frac{2}{3}}}}dx=dt \\ \Rightarrow \frac{dx}{x^{{\displaystyle \frac{2}{3}}}}=3dt \\ \mathrm{Now},\int \frac{dx}{x^{{\displaystyle \frac{2}{3}}}\sqrt{x^{{\displaystyle \frac{2}{3}}}-2^2}} \\ =3\int \frac{dt}{\sqrt{t^2-2^2}} \\ =3\log \left|t+\sqrt{t^2-2^2}\right|+C \\ =3\log \left|x^{\frac{1}{3}}+\sqrt{x^{\frac{2}{3}}-4}\right|+C \end{gathered}$$