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

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closed as off-topic by Mike Miller, Sanath K. Devalapurkar, Hans Engler, Michael Albanese, user61527 Jun 10 '14 at 2:47

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Substitution. Let $u=t^3-1$. – André Nicolas Jun 10 '14 at 0:09
up vote 3 down vote accepted

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}

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I'm a little confused on the $3/4$ part – user124557 Jun 10 '14 at 0:30
Anti-power rule: $$\int y^{\frac 13} \, du= \frac{y^{\frac 13 + 1}}{\frac 13 + 1} + C = \frac{y^{\frac 43}}{\frac 43} + C = \frac 34 u^{\frac 43} + C $$ – Cookie Jun 10 '14 at 0:31

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 $\int u^3\,du$.

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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.

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