What is the integral of x^2\sqrt[3]{1+x^3} ?

The integral of x^2\sqrt[3]{1+x^3} is 1/4 (1+x^3)^{4/3}+C

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integral of x^2\sqrt[3]{1+x^3}

\frac{1}{4} (1 + x^{3})^{\frac{4}{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}{4} u^{\frac{4}{3}}

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

Substitute back

u = 1 + x^{3}

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of x^2\sqrt[3]{1+x^3} ?
The integral of x^2\sqrt[3]{1+x^3} is 1/4 (1+x^3)^{4/3}+C