What is the integral of 6t^2*\sqrt[3]{t} ?

The integral of 6t^2*\sqrt[3]{t} is 6(1/3 t^3\sqrt[3]{t}-1/30 t^{10/3})+C

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integral of 6t^2*\sqrt[3]{t}

\int 6 t^{2} \cdot 3 t d t

6 (\frac{1}{3} t^{3} 3 t - \frac{1}{30} t^{\frac{10}{3}}) + C

Solution steps

\int 6 t^{2} 3 t d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int t^{2} 3 t d t

Apply Integration By Parts

: \frac{1}{3} t^{3} 3 t - \int \frac{1}{9} t^{\frac{7}{3}} d t

= 6 (\frac{1}{3} t^{3} 3 t - \int \frac{1}{9} t^{\frac{7}{3}} d t)

\int \frac{1}{9} t^{\frac{7}{3}} d t = \frac{1}{30} t^{\frac{10}{3}}

= 6 (\frac{1}{3} t^{3} 3 t - \frac{1}{30} t^{\frac{10}{3}})

Add a constant to the solution

= 6 (\frac{1}{3} t^{3} 3 t - \frac{1}{30} t^{\frac{10}{3}}) + C

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What is the integral of 6t^2*\sqrt[3]{t} ?
The integral of 6t^2*\sqrt[3]{t} is 6(1/3 t^3\sqrt[3]{t}-1/30 t^{10/3})+C