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

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

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

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

\frac{( x ^{2} - x ) ^{5}}{5} + C

Solution steps

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

Apply u-substitution

: \int u^{4} d u

= \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{u ^{5}}{5}

Substitute back

u = x^{2} - x

= \frac{( x ^{2} - x ) ^{5}}{5}

Add a constant to the solution

= \frac{( x ^{2} - x ) ^{5}}{5} + C

\frac{d}{d x} ((x^{2} + 2 x)^{2})

\frac{\partial}{\partial x} (5 x^{8} y^{5} + 2 x^{4} y^{6})

\frac{d}{d x} (12 x^{3} cos (x^{4}))

f (x) = \frac{2}{1 - 8 x}

\int \frac{y ^{2}}{2} d y

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