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

1 Answer

Jun 27, 2016

#int x(x-1)^6 dx = 1/8(x-1)^8+1/7(x-1)^7+C#

Explanation:

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#x(x-1)^6 = (x-1+1)(x-1)^6 = (x-1)^7 + (x-1)^6#

So:

#int x(x-1)^6 dx = 1/8(x-1)^8+1/7(x-1)^7+C#

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