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

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

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integral of-x(\sqrt[3]{2x^2+4})

- \frac{3}{16} (2 x^{2} + 4)^{\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{1}{4} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 2 x^{2} + 4

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

Simplify

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

= - \frac{3}{16} (2 x^{2} + 4)^{\frac{4}{3}}

Add a constant to the solution

= - \frac{3}{16} (2 x^{2} + 4)^{\frac{4}{3}} + C

\frac{\partial}{\partial x} (e)

x \to - 19 + lim (k (x))

4 sin (π \cdot x)

\int_{0}^{1} x^{- 2} d x

4 \frac{d y}{d x} + 36 y = 9

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