What is the integral of (t^2)/(\sqrt[3]{t^3+8)} ?

The integral of (t^2)/(\sqrt[3]{t^3+8)} is 1/2 (t^3+8)^{2/3}+C

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integral of (t^2)/(\sqrt[3]{t^3+8)}

\frac{1}{2} (t^{3} + 8)^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{3} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = t^{3} + 8

= \frac{1}{3} \cdot \frac{3}{2} (t^{3} + 8)^{\frac{2}{3}}

Simplify

\frac{1}{3} \cdot \frac{3}{2} (t^{3} + 8)^{\frac{2}{3}} : \frac{1}{2} (t^{3} + 8)^{\frac{2}{3}}

= \frac{1}{2} (t^{3} + 8)^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{2} (t^{3} + 8)^{\frac{2}{3}} + C

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What is the integral of (t^2)/(\sqrt[3]{t^3+8)} ?
The integral of (t^2)/(\sqrt[3]{t^3+8)} is 1/2 (t^3+8)^{2/3}+C