Question

${\int }_{\frac{1}{3}}^1\frac{{\left(x-x^3\right)}^{{\displaystyle \frac{1}{3}}}}{x^4}dx$

Answer 1

Let I = ${\int }_{\frac{1}{3}}^1\frac{{\left(x-x^3\right)}^{{\displaystyle \frac{1}{3}}}}{x^4}dx$
$$\begin{gathered} ={\int }_{\frac{1}{3}}^1\frac{{\left[x^3\left({\displaystyle \frac{x}{x^3}}-1\right)\right]}^{{\displaystyle \frac{1}{3}}}}{x^4}dx \\ ={\int }_{\frac{1}{3}}^1\frac{x{\left({\displaystyle \frac{1}{x^2}}-1\right)}^{{\displaystyle \frac{1}{3}}}}{x^4}dx \\ ={\int }_{\frac{1}{3}}^1\frac{{\left({\displaystyle \frac{1}{x^2}-1}\right)}^{{\displaystyle \frac{1}{3}}}}{x^3}dx \end{gathered}$$

Put $\left(\frac{1}{x^2}-1\right)=z$

$$\begin{gathered} \therefore -\frac{2}{x^3}dx=dz \\ \Rightarrow \frac{dx}{x^3}=-\frac{dz}{2} \end{gathered}$$

When $x\to \frac{1}{3},z\to 8$

When $x\to 1,z\to 0$

$$\begin{gathered} \therefore I=-\frac{1}{2}{\int }_8^0{z}^{\frac{1}{3}}dz \\ =-\frac{1}{2}\times {\overline{)\frac{z^{{\displaystyle \frac{4}{3}}}}{{\displaystyle \frac{4}{3}}}}}_8^0 \\ =-\frac{3}{8}\left[0-{\left(8\right)}^{\frac{4}{3}}\right] \\ =-\frac{3}{8}\times \left(-16\right) \\ =6 \end{gathered}$$