Question

Integrate the function $(x^3-1{)}^{\frac{1}{3}}{x}^5$

Answer 1

Put $x^3-1=t$
$\therefore 3x^{2}dx=dt$
$\Rightarrow \int (x^3-1{)}^{\frac{1}{3}}{x}^{5}dx=\int (x^3-1{)}^{\frac{1}{3}}{x}^3\cdot x^{2}dx$
$=\int t^{\frac{1}{3}}(t+1)\frac{dt}{3}$
$=\frac{1}{3}\int (t^{\frac{4}{3}}+t^{\frac{1}{3}})dt$
$=\frac{1}{3}[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}]+C$
$=\frac{1}{3}[\frac{3}{7}{t}^{\frac{7}{3}}+\frac{3}{4}{t}^{\frac{4}{3}}]+C$
$=\frac{1}{7}(x^3-1{)}^{\frac{7}{3}}+\frac{1}{4}(x^3-1{)}^{\frac{4}{3}}+C$