What is the integral of (2x^2-4x)^9(x-1) ?

The integral of (2x^2-4x)^9(x-1) is 1/40 (2x^2-4x)^{10}+C

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integral of (2x^2-4x)^9(x-1)

\int (2 x^{2} - 4 x)^{9} (x - 1) d x

\frac{1}{40} (2 x^{2} - 4 x)^{10} + C

Solution steps

\int (2 x^{2} - 4 x)^{9} (x - 1) d x

Apply u-substitution

: \int \frac{u ^{9}}{4} d u

= \int \frac{u ^{9}}{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^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= \frac{1}{4} \cdot \frac{u ^{10}}{10}

Substitute back

u = 2 x^{2} - 4 x

= \frac{1}{4} \cdot \frac{( 2 x ^{2} - 4 x ) ^{10}}{10}

Simplify

\frac{1}{4} \cdot \frac{( 2 x ^{2} - 4 x ) ^{10}}{10} : \frac{1}{40} (2 x^{2} - 4 x)^{10}

= \frac{1}{40} (2 x^{2} - 4 x)^{10}

Add a constant to the solution

= \frac{1}{40} (2 x^{2} - 4 x)^{10} + C

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\frac{\partial}{\partial x} (\frac{1}{4 x})

What is the integral of (2x^2-4x)^9(x-1) ?
The integral of (2x^2-4x)^9(x-1) is 1/40 (2x^2-4x)^{10}+C