Integrals — Class 12 Maths Solution

: Let $I = \int {{{\left( {{x^3} - 1} \right)}^{1/3}}{x^5}} dx$

Put ${x^3} - 1 = t$ $\Rightarrow$ $3{x^2}dx = dt$

Also ${x^3} = t + 1$
$\therefore$ $I = \cfrac{1}{3}\int {{{\left( {{x^3} - 1} \right)}^{1/3}}{x^3}} \cdot 3{x^2}dx = \cfrac{1}{3}\int {{t^{1/3}}\left( {t + 1} \right)dt}$

$= \cfrac{1}{3}\left[ {\cfrac{3}{7}{t^{7/3}} + \cfrac{3}{4}{t^{4/3}}} \right] + C = \cfrac{1}{7}{t^{7/3}} + \cfrac{1}{4}{t^{4/3}} + C$

$= \cfrac{1}{7}{\left( {{x^3} - 1} \right)^{7/3}} + \cfrac{1}{4}{\left( {{x^3} - 1} \right)^{4/3}} + C$