What is the integral of x(1-x)^{10} ?

The integral of x(1-x)^{10} is 1/12 (1-x)^{12}-1/11 (1-x)^{11}+C

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integral of x(1-x)^{10}

\int x (1 - x)^{10} d x

\frac{1}{12} (1 - x)^{12} - \frac{1}{11} (1 - x)^{11} + C

Solution steps

\int x (1 - x)^{10} d x

Apply u-substitution

= \int - (- u + 1) u^{10} d u

Take the constant out:

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

= - \int (- u + 1) u^{10} d u

Expand

(- u + 1) u^{10} : - u^{11} + u^{10}

= - \int - u^{11} + u^{10} 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^{11} d u + \int u^{10} d u)

\int u^{11} d u = \frac{u ^{12}}{12}

\int u^{10} d u = \frac{u ^{11}}{11}

= - (- \frac{u ^{12}}{12} + \frac{u ^{11}}{11})

Substitute back

u = 1 - x

= - (- \frac{( 1 - x ) ^{12}}{12} + \frac{( 1 - x ) ^{11}}{11})

Simplify

- (- \frac{( 1 - x ) ^{12}}{12} + \frac{( 1 - x ) ^{11}}{11}) : \frac{1}{12} (1 - x)^{12} - \frac{1}{11} (1 - x)^{11}

= \frac{1}{12} (1 - x)^{12} - \frac{1}{11} (1 - x)^{11}

Add a constant to the solution

= \frac{1}{12} (1 - x)^{12} - \frac{1}{11} (1 - x)^{11} + C

What is the integral of x(1-x)^{10} ?
The integral of x(1-x)^{10} is 1/12 (1-x)^{12}-1/11 (1-x)^{11}+C