Question

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

Answer 1

Consider the problem

$\begin{array}{c}{\int }_{\frac{1}{3}}^1\frac{{\left(x-x^3\right)}^{\frac{1}{3}}}{x^4}\\ ={\int }_{\frac{1}{3}}^1\frac{x{\left(x^{\frac{1}{2}-1}\right)}^{\frac{1}{3}}}{x^4}\\ ={\int }_{\frac{1}{3}}^1{x}^{\frac{1}{3}}{\left(\frac{1}{x^2}-1\right)}^{\frac{1}{3}}dx\end{array}$

putting $\frac{1}{x^2}-1=t$

$\begin{array}{c}\\ \Rightarrow \frac{-2}{x^3}dx=dt\\ \Rightarrow \frac{dx}{x^3}=-\frac{dt}{2}\end{array}$

And

$\begin{array}{c}={\int }_8^0{t}^{\frac{1}{3}}\times \frac{-dt}{2}\\ =\frac{1}{2}{\int }_0^8{t}^{\frac{1}{3}}dt\left[\because {\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx\right]\\ =\frac{1}{2}\times \frac{3}{4}{{\left[t^{\frac{4}{3}}\right]}_0}^8\\ =\frac{3}{8}\left[8^{\frac{4}{3}}\right]\\ =3\times 8^{\frac{1}{3}}\\ =3\times 2\\ =6\end{array}$