Question

The value of the integral ${\int }_{\frac{1}{3}}^1\frac{(x-x^3{)}^{\frac{1}{3}}}{x^4}dx$ is?

• A. $6$
• B. $0$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$6$

$$\begin{array}{c}{\int }_{\frac{1}{3}}^1\frac{{(x-x^3)}^{\frac{1}{3}}}{x^4}dx\\ ={\int }_{\frac{1}{3}}^1\frac{x^3\times \frac{1}{3}{(\frac{1}{x^2}-1)}^{\frac{1}{3}}}{x^4}dx\\ ={\int }_{\frac{1}{3}}^{1}x\frac{{\left(\frac{1}{x^2}-1\right)}^{\frac{1}{3}}}{x^4}dx\\ ={\int }_{\frac{1}{3}}^1\frac{1}{x^3}\left(\frac{1}{x^2}-1\right)dx\\ let{\left(\frac{1}{x^2-1}\right)}^{\frac{1}{3}}=t\\ \frac{1}{3}{\left(\frac{1}{x^2}-1\right)}^{\frac{-2}{3}}\cdot \left(-\frac{1}{x^3}\right)dx=d\\ =-3{\int }_{\frac{1}{3}}^1\frac{1}{x^3}t\cdot x^3{\left(\frac{1}{x^2}-1\right)}^{\frac{2}{3}}dt\\ =-3{\int }_{\frac{1}{3}}^{1}t\cdot t^{2}dt\\ =-3{\int }_{\frac{1}{3}}^1{t}^{3}dt\\ =-3{\left[\frac{t^4}{4}\right]}_{\frac{1}{3}}^1\\ =-3{\left[{\left(\frac{1}{x^2-1}\right)}^{\frac{1}{3}}\right]}_{\frac{1}{3}}^1\\ =3\left[0-2\right]\\ =6\end{array}$$