What is the integral of t\sqrt[3]{t-4} ?

The integral of t\sqrt[3]{t-4} is 3/7 (t-4)^{7/3}+3(t-4)^{4/3}+C

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integral of t\sqrt[3]{t-4}

\frac{3}{7} (t - 4)^{\frac{7}{3}} + 3 (t - 4)^{\frac{4}{3}} + C

Solution steps

Apply u-substitution

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

= \frac{3}{7} u^{\frac{7}{3}} + 3 u^{\frac{4}{3}}

Substitute back

u = t - 4

= \frac{3}{7} (t - 4)^{\frac{7}{3}} + 3 (t - 4)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} (t - 4)^{\frac{7}{3}} + 3 (t - 4)^{\frac{4}{3}} + C

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What is the integral of t\sqrt[3]{t-4} ?
The integral of t\sqrt[3]{t-4} is 3/7 (t-4)^{7/3}+3(t-4)^{4/3}+C