What is the integral of (16x\sqrt[3]{x^2+10})/3 ?

The integral of (16x\sqrt[3]{x^2+10})/3 is 2(x^2+10)^{4/3}+C

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integral of (16x\sqrt[3]{x^2+10})/3

2 (x^{2} + 10)^{\frac{4}{3}} + C

Solution steps

Take the constant out:

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

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{16}{3} \cdot \frac{1}{2} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = x^{2} + 10

= \frac{16}{3} \cdot \frac{1}{2} \cdot \frac{3}{4} (x^{2} + 10)^{\frac{4}{3}}

Simplify

\frac{16}{3} \cdot \frac{1}{2} \cdot \frac{3}{4} (x^{2} + 10)^{\frac{4}{3}} : 2 (x^{2} + 10)^{\frac{4}{3}}

= 2 (x^{2} + 10)^{\frac{4}{3}}

Add a constant to the solution

= 2 (x^{2} + 10)^{\frac{4}{3}} + C

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What is the integral of (16x\sqrt[3]{x^2+10})/3 ?
The integral of (16x\sqrt[3]{x^2+10})/3 is 2(x^2+10)^{4/3}+C