What is the integral of x\sqrt[3]{1-x} ?

The integral of x\sqrt[3]{1-x} is 3/7 (1-x)^{7/3}-3/4 (1-x)^{4/3}+C

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integral of x\sqrt[3]{1-x}

\int x 3 1 - x d x

\frac{3}{7} (1 - x)^{\frac{7}{3}} - \frac{3}{4} (1 - x)^{\frac{4}{3}} + C

Solution steps

\int x 3 1 - x d x

Apply u-substitution

: \int - (- u + 1) 3 u d u

= \int - (- u + 1) 3 u d u

Take the constant out:

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

= - \int (- u + 1) 3 u d u

Expand

(- u + 1) 3 u : - u^{\frac{4}{3}} + 3 u

= - \int - u^{\frac{4}{3}} + 3 u d u

Apply the Sum Rule:

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

= - (- \int u^{\frac{4}{3}} d u + \int 3 u d u)

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

\int 3 u d u = \frac{3}{4} u^{\frac{4}{3}}

= - (- \frac{3}{7} u^{\frac{7}{3}} + \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 1 - x

= - (- \frac{3}{7} (1 - x)^{\frac{7}{3}} + \frac{3}{4} (1 - x)^{\frac{4}{3}})

Simplify

= \frac{3}{7} (1 - x)^{\frac{7}{3}} - \frac{3}{4} (1 - x)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} (1 - x)^{\frac{7}{3}} - \frac{3}{4} (1 - x)^{\frac{4}{3}} + C

\int x ln (\frac{x}{2}) d x

\frac{d}{d x} (x + 5 - 3)

\int e^{x} \cdot cos (4 x) d x

x \to \infty lim (2 x - 4)

x - 2 = - (\frac{1}{2}) (y - 4)^{2}, x + y = 6

What is the integral of x\sqrt[3]{1-x} ?
The integral of x\sqrt[3]{1-x} is 3/7 (1-x)^{7/3}-3/4 (1-x)^{4/3}+C