What is the integral of \sqrt[3]{3t+8} ?

The integral of \sqrt[3]{3t+8} is 1/4 (3t+8)^{4/3}+C

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integral of \sqrt[3]{3t+8}

\int 3 3 t + 8 d t

\frac{1}{4} (3 t + 8)^{\frac{4}{3}} + C

Solution steps

\int 3 3 t + 8 d t

Apply u-substitution

: \int \frac{3 u}{3} d u

= \int \frac{3 u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int 3 u d u

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{3} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 3 t + 8

= \frac{1}{3} \cdot \frac{3}{4} (3 t + 8)^{\frac{4}{3}}

Simplify

\frac{1}{3} \cdot \frac{3}{4} (3 t + 8)^{\frac{4}{3}} : \frac{1}{4} (3 t + 8)^{\frac{4}{3}}

= \frac{1}{4} (3 t + 8)^{\frac{4}{3}}

Add a constant to the solution

= \frac{1}{4} (3 t + 8)^{\frac{4}{3}} + C

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What is the integral of \sqrt[3]{3t+8} ?
The integral of \sqrt[3]{3t+8} is 1/4 (3t+8)^{4/3}+C