What is the integral of 6t^4\sqrt[3]{t} ?

The integral of 6t^4\sqrt[3]{t} is 9/8 t^{16/3}+C

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integral of 6t^4\sqrt[3]{t}

\int 6 t^{4} 3 t d t

\frac{9}{8} t^{\frac{16}{3}} + C

Solution steps

\int 6 t^{4} 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^{4} 3 t d t

Simplify

t^{4} 3 t : t^{\frac{13}{3}}

= 6 \cdot \int t^{\frac{13}{3}} d t

Apply the Power Rule

: \frac{3}{16} t^{\frac{16}{3}}

= 6 \cdot \frac{3}{16} t^{\frac{16}{3}}

Simplify

6 \cdot \frac{3}{16} t^{\frac{16}{3}} : \frac{9}{8} t^{\frac{16}{3}}

= \frac{9}{8} t^{\frac{16}{3}}

Add a constant to the solution

= \frac{9}{8} t^{\frac{16}{3}} + C

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What is the integral of 6t^4\sqrt[3]{t} ?
The integral of 6t^4\sqrt[3]{t} is 9/8 t^{16/3}+C