What is the integral of x(1-2x)^{99} ?

The integral of x(1-2x)^{99} is 1/4 (((1-2x)^{101})/(101)-((1-2x)^{100})/(100))+C

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integral of x(1-2x)^{99}

\int x (1 - 2 x)^{99} d x

\frac{1}{4} (\frac{( 1 - 2 x ) ^{101}}{101} - \frac{( 1 - 2 x ) ^{100}}{100}) + C

Solution steps

\int x (1 - 2 x)^{99} d x

Apply u-substitution

: \int \frac{u ^{99} ( u - 1 )}{4} d u

= \int \frac{u ^{99} ( u - 1 )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{99} (u - 1) d u

Expand

u^{99} (u - 1) : u^{100} - u^{99}

= \frac{1}{4} \cdot \int u^{100} - u^{99} 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}{4} (\int u^{100} d u - \int u^{99} d u)

\int u^{100} d u = \frac{u ^{101}}{101}

\int u^{99} d u = \frac{u ^{100}}{100}

= \frac{1}{4} (\frac{u ^{101}}{101} - \frac{u ^{100}}{100})

Substitute back

u = 1 - 2 x

= \frac{1}{4} (\frac{( 1 - 2 x ) ^{101}}{101} - \frac{( 1 - 2 x ) ^{100}}{100})

Add a constant to the solution

= \frac{1}{4} (\frac{( 1 - 2 x ) ^{101}}{101} - \frac{( 1 - 2 x ) ^{100}}{100}) + C

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What is the integral of x(1-2x)^{99} ?
The integral of x(1-2x)^{99} is 1/4 (((1-2x)^{101})/(101)-((1-2x)^{100})/(100))+C