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

The integral of (x-x^2)/(2\sqrt[3]{x)} is 1/2 (3/5 x^{5/3}-3/8 x^{8/3})+C

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

\int \frac{x - x ^{2}}{2 3 x} d x

\frac{1}{2} (\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{8} x^{\frac{8}{3}}) + C

Solution steps

\int \frac{x - x ^{2}}{2 3 x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{x - x ^{2}}{3 x} d x

Expand

\frac{x - x ^{2}}{3 x} : \frac{x}{3 x} - \frac{x ^{2}}{3 x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{x}{3 x} d x - \int \frac{x ^{2}}{3 x} d x)

\int \frac{x}{3 x} d x = \frac{3}{5} x^{\frac{5}{3}}

\int \frac{x ^{2}}{3 x} d x = \frac{3}{8} x^{\frac{8}{3}}

= \frac{1}{2} (\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{8} x^{\frac{8}{3}})

Add a constant to the solution

= \frac{1}{2} (\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{8} x^{\frac{8}{3}}) + C

\frac{\partial}{\partial x} (- 5 e^{x} ln (y))

y = x^{2}, y = 3 x + 4

- 6

\frac{d}{d t} (e^{t} - 4)

(2 x - y) + (2 y - x) y^{^{'}} = 0, y (1) = 3

What is the integral of (x-x^2)/(2\sqrt[3]{x)} ?
The integral of (x-x^2)/(2\sqrt[3]{x)} is 1/2 (3/5 x^{5/3}-3/8 x^{8/3})+C