What is the integral of t\sqrt[3]{t+10} ?

The integral of t\sqrt[3]{t+10} is 3/7 (t+10)^{7/3}-15/2 (t+10)^{4/3}+C

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integral of t\sqrt[3]{t+10}

\frac{3}{7} (t + 10)^{\frac{7}{3}} - \frac{15}{2} (t + 10)^{\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}} - \frac{15}{2} u^{\frac{4}{3}}

Substitute back

u = t + 10

= \frac{3}{7} (t + 10)^{\frac{7}{3}} - \frac{15}{2} (t + 10)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} (t + 10)^{\frac{7}{3}} - \frac{15}{2} (t + 10)^{\frac{4}{3}} + C

What is the integral of t\sqrt[3]{t+10} ?
The integral of t\sqrt[3]{t+10} is 3/7 (t+10)^{7/3}-15/2 (t+10)^{4/3}+C