How do you find antiderivative of #(1-x)^2#? Calculus Introduction to Integration Integrals of Polynomial functions 1 Answer Jim S May 3, 2018 #(x-1)^3/3+c# Explanation: #int(1-x)^2dx=# Substitute #1-x=u# #-dx=du# #dx=-du# #intu^2(-du)# #=# #-intu^2du# #=# #-int(u^3/3)'du# #=# #-u^3/3+c# #=# #(x-1)^3/3+c# , #c##in##RR# Answer link Related questions How do you evaluate the integral #intx^3+4x^2+5 dx#? How do you evaluate the integral #int(1+x)^2 dx#? How do you evaluate the integral #int8x+3 dx#? How do you evaluate the integral #intx^10-6x^5+2x^3 dx#? What is the integral of a constant? What is the antiderivative of the distance function? What is the integral of #|x|#? What is the integral of #3x#? What is the integral of #4x^3#? What is the integral of #sqrt(1-x^2)#? See all questions in Integrals of Polynomial functions Impact of this question 1521 views around the world You can reuse this answer Creative Commons License