How do you integrate $x {(x - 1)}^{6} d x$ $x(x-1)^6 dx$ ?

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Answer 1 · 2016-06-27T06:14:21

$\int x {(x - 1)}^{6} d x = \frac{1}{8} {(x - 1)}^{8} + \frac{1}{7} {(x - 1)}^{7} + C$ $int x(x-1)^6 dx = 1/8(x-1)^8+1/7(x-1)^7+C$

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$x {(x - 1)}^{6} = (x - 1 + 1) {(x - 1)}^{6} = {(x - 1)}^{7} + {(x - 1)}^{6}$ $x(x-1)^6 = (x-1+1)(x-1)^6 = (x-1)^7 + (x-1)^6$

So:

$\int x {(x - 1)}^{6} d x = \frac{1}{8} {(x - 1)}^{8} + \frac{1}{7} {(x - 1)}^{7} + C$ $int x(x-1)^6 dx = 1/8(x-1)^8+1/7(x-1)^7+C$

How do you integrate x(x−1)6dxx(x-1)^6 dx?