How do you integrate #x(x-1)^6 dx#?
1 Answer
Jun 27, 2016
Explanation:
#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#
#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#