I rewrote this as
= t(t - 4)^1/3.... do I combine them and integrate after?
$\displaystyle \text{We know that } \int u \, dv=uv-\int v \, du.\!$
So for $\displaystyle \int t\sqrt[3]{t-4} \;\;dt$ let $\displaystyle u = t \therefore du = dt$
And let $\displaystyle dv = \sqrt[3]{t-4} \;\;dt \therefore v = \frac{3}{4}(t - 4)^{\frac{4}{3}}$
$\displaystyle \begin{align*}\int t\sqrt[3]{t-4} \;\;dt =& \frac{3}{4}t(t-4)^{\frac{4}{3}} - \int \frac{3}{4}(t - 4)^{\frac{4}{3}} \\ =& \frac{3}{4}t(t-4)^{\frac{4}{3}} - \frac{9}{28}(t-4)^{\frac{7}{3}} + C \\ =& \frac{3}{7}(t-4)^{\frac{4}{3}}[\frac{7t}{4} - \frac{3t}{4} + \frac{12}{4}] + C \\ =& \frac{3}{7}(t-4)^{\frac{4}{3}}(t+3) + C\end{align*}$
$\displaystyle \therefore \int t\sqrt[3]{t-4} \;\;dt = \frac{3}{7}(t-4)^{\frac{4}{3}}(t+3) + C$
Check:
int x(x-4)^(1/3) dx - Wolfram|Alpha
Or starting from skeeter's post,
$\displaystyle \int (u+4) \cdot \sqrt[3]{u} \, du =\int u^\frac{4}{3} + 4u^\frac{1}{3} \, du =$
$\displaystyle \frac{3}{7}u^\frac{7}{3}+ 3 u^\frac{4}{3} + C = \frac{3}{7}u^\frac{4}{3} (u + 7) + C=$
$\displaystyle \frac{3}{7} (t-4)^\frac{4}{3} (t+3) + C$
- Hollywood
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