What is the integral of x(1-2x)^4 ?

The integral of x(1-2x)^4 is (x^2)/2-(8x^3)/3+6x^4-(32x^5)/5+(8x^6)/3+C

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integral of x(1-2x)^4

\int x (1 - 2 x)^{4} d x

\frac{x ^{2}}{2} - \frac{8 x ^{3}}{3} + 6 x^{4} - \frac{32 x ^{5}}{5} + \frac{8 x ^{6}}{3} + C

Solution steps

\int x (1 - 2 x)^{4} d x

Factor

1 - 2 x : - (- 1 + 2 x)

= \int x (- (- 1 + 2 x))^{4} d x

(- (- 1 + 2 x))^{4} = (- 1 + 2 x)^{4}

= \int x (- 1 + 2 x)^{4} d x

Expand

x (- 1 + 2 x)^{4} : x - 8 x^{2} + 24 x^{3} - 32 x^{4} + 16 x^{5}

= \int x - 8 x^{2} + 24 x^{3} - 32 x^{4} + 16 x^{5} d x

Apply the Sum Rule:

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

= \int x d x - \int 8 x^{2} d x + \int 24 x^{3} d x - \int 32 x^{4} d x + \int 16 x^{5} d x

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

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

\int 24 x^{3} d x = 6 x^{4}

\int 32 x^{4} d x = \frac{32 x ^{5}}{5}

\int 16 x^{5} d x = \frac{8 x ^{6}}{3}

= \frac{x ^{2}}{2} - \frac{8 x ^{3}}{3} + 6 x^{4} - \frac{32 x ^{5}}{5} + \frac{8 x ^{6}}{3}

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{8 x ^{3}}{3} + 6 x^{4} - \frac{32 x ^{5}}{5} + \frac{8 x ^{6}}{3} + C

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What is the integral of x(1-2x)^4 ?
The integral of x(1-2x)^4 is (x^2)/2-(8x^3)/3+6x^4-(32x^5)/5+(8x^6)/3+C