What is the integral of \sqrt[3]{x^4+3}(x^3) ?

The integral of \sqrt[3]{x^4+3}(x^3) is 3/16 (x^4+3)^{4/3}+C

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integral of \sqrt[3]{x^4+3}(x^3)

\frac{3}{16} (x^{4} + 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 u-substitution

Apply the Power Rule

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

Substitute back

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

Simplify

\frac{1}{4} \cdot \frac{3}{4} (x^{4} + 3)^{\frac{4}{3}} : \frac{3}{16} (x^{4} + 3)^{\frac{4}{3}}

= \frac{3}{16} (x^{4} + 3)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{16} (x^{4} + 3)^{\frac{4}{3}} + C

What is the integral of \sqrt[3]{x^4+3}(x^3) ?
The integral of \sqrt[3]{x^4+3}(x^3) is 3/16 (x^4+3)^{4/3}+C