∫f′(ax+b)[f(ax+b)]n dx.
Let I=∫f′(ax+b)[f(ax+b)]n dxPut f(ax+b)=t⇒f′(ax+b) a dx=dt=dtaf′(ax+b)∴ I=∫f′(ax+b)tndtf′(ax+b)a=1a∫tndt=1a(tn+1n+1)+C=1a.[f(ax+b)]n+1n+1+C