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Question

f(ax+b)[f(ax+b)]n dx.

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Solution

Let I=f(ax+b)[f(ax+b)]n dxPut f(ax+b)=tf(ax+b) a dx=dt=dtaf(ax+b) I=f(ax+b)tndtf(ax+b)a=1atndt=1a(tn+1n+1)+C=1a.[f(ax+b)]n+1n+1+C


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