What is the integral of x(2x-1)^{100} ?

The integral of x(2x-1)^{100} is 1/4 (((2x-1)^{102})/(102)+((2x-1)^{101})/(101))+C

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integral of x(2x-1)^{100}

\int x (2 x - 1)^{100} d x

\frac{1}{4} (\frac{( 2 x - 1 ) ^{102}}{102} + \frac{( 2 x - 1 ) ^{101}}{101}) + C

Solution steps

\int x (2 x - 1)^{100} d x

Apply u-substitution

= \int \frac{u ^{100} ( 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^{100} (u + 1) d u

Expand

u^{100} (u + 1) : u^{101} + u^{100}

= \frac{1}{4} \cdot \int u^{101} + u^{100} 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^{101} d u + \int u^{100} d u)

\int u^{101} d u = \frac{u ^{102}}{102}

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

= \frac{1}{4} (\frac{u ^{102}}{102} + \frac{u ^{101}}{101})

Substitute back

u = 2 x - 1

= \frac{1}{4} (\frac{( 2 x - 1 ) ^{102}}{102} + \frac{( 2 x - 1 ) ^{101}}{101})

Add a constant to the solution

= \frac{1}{4} (\frac{( 2 x - 1 ) ^{102}}{102} + \frac{( 2 x - 1 ) ^{101}}{101}) + C

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What is the integral of x(2x-1)^{100} ?
The integral of x(2x-1)^{100} is 1/4 (((2x-1)^{102})/(102)+((2x-1)^{101})/(101))+C