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

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

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integral of x(1-x)^4

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

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

Solution steps

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

Expand

x (1 - x)^{4} : x - 4 x^{2} + 6 x^{3} - 4 x^{4} + x^{5}

= \int x - 4 x^{2} + 6 x^{3} - 4 x^{4} + 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 4 x^{2} d x + \int 6 x^{3} d x - \int 4 x^{4} d x + \int x^{5} d x

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

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

\int x^{5} d x = \frac{x ^{6}}{6}

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

Add a constant to the solution

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

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