What is the integral of 2x\sqrt[3]{64+x^2} ?

The integral of 2x\sqrt[3]{64+x^2} is 3/4 (64+x^2)^{4/3}+C

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integral of 2x\sqrt[3]{64+x^2}

\frac{3}{4} (64 + x^{2})^{\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}}

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

Substitute back

u = 64 + x^{2}

= 2 \cdot \frac{1}{2} \cdot \frac{3}{4} (64 + x^{2})^{\frac{4}{3}}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{3}{4} (64 + x^{2})^{\frac{4}{3}} : \frac{3}{4} (64 + x^{2})^{\frac{4}{3}}

= \frac{3}{4} (64 + x^{2})^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{4} (64 + x^{2})^{\frac{4}{3}} + C

\frac{\partial}{\partial x} (y^{2} - x \cdot y)

\int ln (2 - x) d x

\frac{d}{d x} (ln (x^{3} - 2 x^{2}))

\frac{d ^{2}}{d x ^{2}} (x^{\frac{5}{2}} e^{x})

\int cos (0)

What is the integral of 2x\sqrt[3]{64+x^2} ?
The integral of 2x\sqrt[3]{64+x^2} is 3/4 (64+x^2)^{4/3}+C