How do you find antiderivative of #(1-x)^2#?

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Answer 1 · 2018-05-03T19:51:50

#(x-1)^3/3+c#

#int(1-x)^2dx=#

Substitute #1-x=u#

#-dx=du#
#dx=-du#

#-intu^2du# #=#

#-int(u^3/3)'du# #=#

#-u^3/3+c# #=#

#(x-1)^3/3+c#

, #c##in##RR#