Find the indefinite integral

Question

$\displaystyle\int{t^2}{{\sqrt[3]{t^3-1}}}\,dt$. Can someone give me a hint on how to solve this problem?

Answer 1

Substitute the following

• $u=t^3-1$
• $du=3t^2 \, dt \implies \frac{du}{3}= t^2 \, dt$

So that we may now perform the integration by $u$-substitution. \begin{align}\int t^2 \sqrt[3]{t^3-1} \, dt &= \int \sqrt[3]{t^3-1} \, \underbrace{t^2 \, dt}_{du/3}\\ &= \int \sqrt[3] u \, \frac{du}{3} \\ &=\frac 13 \int u^{\frac 13} \, du \\ &=\frac 13 \frac 34u^{\frac 43} +C \\ &=\frac 14(t^3-1)^{\frac 43} +C\end{align}

Answer 2

Let $u=t^3-1\implies du=3t^2dt$. Hence, $$\dfrac{1}{3}\int \sqrt[3]{t^3-1}(3t^2dt)=\dfrac{1}{3}\int \sqrt[3]{u}du$$ You should be able to do this.

Answer 3

The usual substitution is $u=t^3-1$. Alternately, let $u^3=t^3-1$. Then $3u^2\,du=3t^2\,dt$, so we want $\int u^3\,du$.