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

The integral of (2x-1)(-x^2+x)^4 is (x^{10})/5-x^9+2x^8-2x^7+x^6-(x^5)/5+C

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

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

\frac{x ^{10}}{5} - x^{9} + 2 x^{8} - 2 x^{7} + x^{6} - \frac{x ^{5}}{5} + C

Solution steps

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

Expand

(2 x - 1) (- x^{2} + x)^{4} : 2 x^{9} - 9 x^{8} + 16 x^{7} - 14 x^{6} + 6 x^{5} - x^{4}

= \int 2 x^{9} - 9 x^{8} + 16 x^{7} - 14 x^{6} + 6 x^{5} - x^{4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 x^{9} d x - \int 9 x^{8} d x + \int 16 x^{7} d x - \int 14 x^{6} d x + \int 6 x^{5} d x - \int x^{4} d x

\int 2 x^{9} d x = \frac{x ^{10}}{5}

\int 9 x^{8} d x = x^{9}

\int 16 x^{7} d x = 2 x^{8}

\int 14 x^{6} d x = 2 x^{7}

\int 6 x^{5} d x = x^{6}

\int x^{4} d x = \frac{x ^{5}}{5}

= \frac{x ^{10}}{5} - x^{9} + 2 x^{8} - 2 x^{7} + x^{6} - \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{x ^{10}}{5} - x^{9} + 2 x^{8} - 2 x^{7} + x^{6} - \frac{x ^{5}}{5} + C

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What is the integral of (2x-1)(-x^2+x)^4 ?
The integral of (2x-1)(-x^2+x)^4 is (x^{10})/5-x^9+2x^8-2x^7+x^6-(x^5)/5+C