Question

Integrate the following functions.
$\int (x^3-1{)}^{\frac{1}{3}}{x}^{5}dx.$

Answer 1

$\int (x^3-1{)}^{\frac{1}{3}}{x}^{5}dx=\int (x^3-1{)}^{\frac{1}{3}}{x}^3.x^{2}dx$
Let $x^3-1=t\Rightarrow x^3=t+1$
On differentiating w.r.t.x, we get $3x^2=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{3x^2}$
$\therefore \int (x^3-1{)}^{\frac{1}{3}}{x}^3.x^{2}dx=\int t^{\frac{1}{3}}(t+1)x^2\frac{dt}{3x^2}=\frac{1}{3}\int (t^{\frac{4}{3}}+t^{\frac{1}{3}})dt=\frac{1}{3}[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}]+C=\frac{1}{3}[\frac{3}{7}{t}^{\frac{7}{3}}+\frac{3}{4}{t}^{\frac{4}{3}}]+C=\frac{1}{7}{t}^{\frac{7}{3}}+\frac{1}{4}{t}^{\frac{4}{3}}+C=\frac{1}{7}(x^3-1{)}^{\frac{7}{3}}+\frac{1}{4}(x^3-1{)}^{\frac{4}{3}}+C$