Question #0228a Calculus Introduction to Integration Basic Properties of Definite Integrals 1 Answer Ratnaker Mehta Aug 14, 2016 #(1) : int(1-x)^ndx=-(1-x)^(n+1)/(n+1)+C, if, n!=-1# #(2) : int(1-x)^ndx=-ln|1-x|+C', or, ln|1/(1-x)|+C', if n=-1#. Explanation: Let #I=int(1-x)^ndx# Take, #(1-x)=t rArr -dx=dt# Hence, #I=int t^n(-dt)=-int t^ndt# #=-t^(n+1)/(n+1), if, n+1!=0, i.e., n!=-1# #=-(1-x)^(n+1)/(n+1)+C, if, n!=-1# In case, #n=-1, #I=intt^-1(-dt)=-int1/tdt=-ln|t|=-ln|1-x|+C'#, or, #=ln|1/(1-x)|+C'# Answer link Related questions What are five basic properties of definite integrals? What is the integral of an integral? What is the integral of a quotient? How do I evaluate #int_0^5(2 e^x + 5cos(x)) dx#? Why can't you integrate #sqrt(1+(cosx/-sinx)^2#? What are the different strategies of integration? What is the integral from 0 to 4 of lnx dx? How do you find the integral of 0 to the infinity of #x^(8/3) dx#? How do you evaluate the integral of #(ln x)^2 dx#? How do you find the integral of #abs(x) dx# on the interval [-2, 1]? See all questions in Basic Properties of Definite Integrals Impact of this question 3918 views around the world You can reuse this answer Creative Commons License