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

The integral of x\sqrt[3]{1-3x} is 1/84 (4(-3x+1)^{7/3}-7(-3x+1)^{4/3})+C

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

\int x 3 1 - 3 x d x

\frac{1}{84} (4 (- 3 x + 1)^{\frac{7}{3}} - 7 (- 3 x + 1)^{\frac{4}{3}}) + C

Solution steps

\int x 3 1 - 3 x d x

Apply u-substitution

: \int \frac{3 u ( u - 1 )}{9} d u

= \int \frac{3 u ( u - 1 )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int 3 u (u - 1) d u

Expand

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

= \frac{1}{9} \cdot \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

= \frac{1}{9} (\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{1}{9} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 1 - 3 x

= \frac{1}{9} (\frac{3}{7} (1 - 3 x)^{\frac{7}{3}} - \frac{3}{4} (1 - 3 x)^{\frac{4}{3}})

Simplify

\frac{1}{9} (\frac{3}{7} (1 - 3 x)^{\frac{7}{3}} - \frac{3}{4} (1 - 3 x)^{\frac{4}{3}}) : \frac{1}{84} (4 (- 3 x + 1)^{\frac{7}{3}} - 7 (- 3 x + 1)^{\frac{4}{3}})

= \frac{1}{84} (4 (- 3 x + 1)^{\frac{7}{3}} - 7 (- 3 x + 1)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{84} (4 (- 3 x + 1)^{\frac{7}{3}} - 7 (- 3 x + 1)^{\frac{4}{3}}) + C

\int (6 x^{2} + 4 x^{\frac{1}{3}}) d x

\int 7 csc^{2} (7 x) cot (7 x) d x

\int \frac{20}{( x - 1 ) x ^{2} - 2 x - 48} d x

\int 2 x^{2} 4 - x^{2} d x

\int x (50 - x) d x

What is the integral of x\sqrt[3]{1-3x} ?
The integral of x\sqrt[3]{1-3x} is 1/84 (4(-3x+1)^{7/3}-7(-3x+1)^{4/3})+C