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How do you integrate #int(3-x)^10 dx#?

1 Answer

Feb 23, 2015

Hello,

Consider #F(x) = - (3-x)^(11)/11#. So, #F'(x) = (3-x)^(10)#. Therefore,

#int (3-x)^(10) dx = - (3-x)^(11)/11 + c#

where #c in RR# is an arbitrary constant.

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