f'(ax+b)[f(ax+b)]^(n)...

`f'(ax+b)[f(ax+b)]^(n)`

Verified by Experts

माना `f(ax+b)=t rArr f'(ax+b)adx=dt`
`therefore" "I=(1)/(a)t^(n)dt=(1)/(a)((t^(n+1))/(n+1))+C`
`" "=(1)/(a).([f(ax+b)]^(n+1))/(n+1)+C`

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