What is the integral of x(2x-1)^7 ?

The integral of x(2x-1)^7 is (128x^9)/9-56x^8+96x^7-(280x^6)/3+56x^5-21x^4+(14x^3)/3-(x^2)/2+C

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integral of x(2x-1)^7

\int x (2 x - 1)^{7} d x

\frac{128 x ^{9}}{9} - 56 x^{8} + 96 x^{7} - \frac{280 x ^{6}}{3} + 56 x^{5} - 21 x^{4} + \frac{14 x ^{3}}{3} - \frac{x ^{2}}{2} + C

Solution steps

\int x (2 x - 1)^{7} d x

Expand

x (2 x - 1)^{7} : 128 x^{8} - 448 x^{7} + 672 x^{6} - 560 x^{5} + 280 x^{4} - 84 x^{3} + 14 x^{2} - x

= \int 128 x^{8} - 448 x^{7} + 672 x^{6} - 560 x^{5} + 280 x^{4} - 84 x^{3} + 14 x^{2} - x 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 128 x^{8} d x - \int 448 x^{7} d x + \int 672 x^{6} d x - \int 560 x^{5} d x + \int 280 x^{4} d x - \int 84 x^{3} d x + \int 14 x^{2} d x - \int x d x

\int 128 x^{8} d x = \frac{128 x ^{9}}{9}

\int 448 x^{7} d x = 56 x^{8}

\int 672 x^{6} d x = 96 x^{7}

\int 560 x^{5} d x = \frac{280 x ^{6}}{3}

\int 280 x^{4} d x = 56 x^{5}

\int 84 x^{3} d x = 21 x^{4}

\int 14 x^{2} d x = \frac{14 x ^{3}}{3}

\int x d x = \frac{x ^{2}}{2}

= \frac{128 x ^{9}}{9} - 56 x^{8} + 96 x^{7} - \frac{280 x ^{6}}{3} + 56 x^{5} - 21 x^{4} + \frac{14 x ^{3}}{3} - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{128 x ^{9}}{9} - 56 x^{8} + 96 x^{7} - \frac{280 x ^{6}}{3} + 56 x^{5} - 21 x^{4} + \frac{14 x ^{3}}{3} - \frac{x ^{2}}{2} + C

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What is the integral of x(2x-1)^7 ?
The integral of x(2x-1)^7 is (128x^9)/9-56x^8+96x^7-(280x^6)/3+56x^5-21x^4+(14x^3)/3-(x^2)/2+C