What is the integral of x^3sqrt(1-2x) ?

The integral of x^3sqrt(1-2x) is -1/315 (35x^3+15x^2+6x+2)(-2x+1)^{3/2}+C

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integral of x^3sqrt(1-2x)

\int x^{3} 1 - 2 x d x

- \frac{1}{315} (35 x^{3} + 15 x^{2} + 6 x + 2) (- 2 x + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{3} 1 - 2 x d x

Apply u-substitution

: \int \frac{u ( u - 1 ) ^{3}}{16} d u

= \int \frac{u ( u - 1 ) ^{3}}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u (u - 1)^{3} d u

Expand

u (u - 1)^{3} : u^{\frac{7}{2}} - 3 u^{\frac{5}{2}} + 3 u^{\frac{3}{2}} - u

= \frac{1}{16} \cdot \int u^{\frac{7}{2}} - 3 u^{\frac{5}{2}} + 3 u^{\frac{3}{2}} - 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}{16} (\int u^{\frac{7}{2}} d u - \int 3 u^{\frac{5}{2}} d u + \int 3 u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{7}{2}} d u = \frac{2}{9} u^{\frac{9}{2}}

\int 3 u^{\frac{5}{2}} d u = \frac{6}{7} u^{\frac{7}{2}}

\int 3 u^{\frac{3}{2}} d u = \frac{6}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{16} (\frac{2}{9} u^{\frac{9}{2}} - \frac{6}{7} u^{\frac{7}{2}} + \frac{6}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - 2 x

= \frac{1}{16} (\frac{2}{9} (1 - 2 x)^{\frac{9}{2}} - \frac{6}{7} (1 - 2 x)^{\frac{7}{2}} + \frac{6}{5} (1 - 2 x)^{\frac{5}{2}} - \frac{2}{3} (1 - 2 x)^{\frac{3}{2}})

Simplify

\frac{1}{16} (\frac{2}{9} (1 - 2 x)^{\frac{9}{2}} - \frac{6}{7} (1 - 2 x)^{\frac{7}{2}} + \frac{6}{5} (1 - 2 x)^{\frac{5}{2}} - \frac{2}{3} (1 - 2 x)^{\frac{3}{2}}) : - \frac{1}{315} (35 x^{3} + 15 x^{2} + 6 x + 2) (- 2 x + 1)^{\frac{3}{2}}

= - \frac{1}{315} (35 x^{3} + 15 x^{2} + 6 x + 2) (- 2 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{315} (35 x^{3} + 15 x^{2} + 6 x + 2) (- 2 x + 1)^{\frac{3}{2}} + C

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What is the integral of x^3sqrt(1-2x) ?
The integral of x^3sqrt(1-2x) is -1/315 (35x^3+15x^2+6x+2)(-2x+1)^{3/2}+C