Question

please answer this question​

Answer 1

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

can be written as:-

$\rm\int{\frac{- \frac{x^{3}}{3} + \frac{x}{3}}{x^{4}} d x}$

$\rm = {\int{\frac{1 - x^{2}}{3 x^{3}} d x}}$

$\rm = {\left(\frac{\displaystyle \rm\int{\frac{1 - x^{2}}{x^{3}} d x}}{3}\right)}$

$= \rm \frac{{\displaystyle \rm\int{\left(- \frac{1}{x} + \frac{1}{x^{3}}\right)d x}}}{3}$

$= \rm \frac{{\left( \displaystyle \rm\int{\frac{1}{x^{3}} d x} - \int{\frac{1}{x} d x}\right)}}{3}$

$\rm \red{- \frac{\int{\frac{1}{x} d x}}{3} + \frac{\color{red}{\int{\frac{1}{x^{3}} d x}}}{3}=- \frac{\int{\frac{1}{x} d x}}{3} + \frac{\color{red}{\int{x^{-3} d x}}}{3}=- \frac{\int{\frac{1}{x} d x}}{3} + \frac{\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}{3}=- \frac{\int{\frac{1}{x} d x}}{3} + \frac{\color{red}{\left(- \frac{x^{-2}}{2}\right)}}{3}=- \frac{\int{\frac{1}{x} d x}}{3} + \frac{\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}{3}}$

$\rm\int{\frac{- \frac{x^{3}}{3} + \frac{x}{3}}{x^{4}} d x} = - \frac{\ln{\left(\left|{x}\right| \right)}}{3} - \frac{1}{6 x^{2}}+C$

Answer 2

Simplify the integrand:

$\dfrac{\left(x-x^3\right)^{\frac13}}{x^4} = \dfrac{\left(x^3 \left(\frac1{x^2}-1\right)\right)^{\frac13}}{x^4}= \dfrac{x \left(\frac1{x^2} - 1\right)^{\frac13}}{x^4} = \dfrac{\left(\frac1{x^2}-1\right)^{\frac13}}{x^3}$

Substitute y = 1/x² - 1 and dy = -2/x³ dx :

$\displaystyle \int \frac{\left(\frac1{x^2}-1\right)^{\frac13}}{x^3} \, dx = -\frac12 \int y^{\frac13} \, dy \\\\ = -\frac12 \times \frac34 y^{\frac43} + C \\\\ = \boxed{-\frac38 \left(\frac1{x^2}-1\right)^{\frac43} + C}$