Question

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

Answer 1

Let x3 - 1 = t

∴ 3x2dx = dt

$\Rightarrow \int(x^3-1)^{\frac{1}{3}}x^5dx= \int x^3.x^2dx$

= $\int t^{\frac{1}{3}}(t+1)\frac{dt}{3}$

=$\frac{1}{3}\int \bigg(t^{\frac{4}{3}}+t^{\frac{1}{3}}\big)dt$

= $\frac{1}{3}\bigg[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\bigg]+C$

= $\frac{1}{3}\bigg[\frac{3}{7}t^{\frac{7}{3}}+\frac{3}{4}t^{\frac{4}{3}}\bigg]+C$

= $\frac{1}{7}(x^3-1)^{\frac{7}{3}}+\frac{1}{4}(x^3-1)^{\frac{4}{3}}+C$