What is the integral of (t^2)/(\sqrt[3]{6+t^3)} ?

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

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integral of (t^2)/(\sqrt[3]{6+t^3)}

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

\frac{1}{2} (6 + t^{3})^{\frac{2}{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 3 u} d u

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

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 = 6 + t^{3}

= \frac{1}{3} \cdot \frac{3}{2} (6 + t^{3})^{\frac{2}{3}}

Simplify

\frac{1}{3} \cdot \frac{3}{2} (6 + t^{3})^{\frac{2}{3}} : \frac{1}{2} (6 + t^{3})^{\frac{2}{3}}

= \frac{1}{2} (6 + t^{3})^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{2} (6 + t^{3})^{\frac{2}{3}} + C

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