Use the reduction formula: [(In 2)" dx = 2(In x)" — n f (In 2)¹-1, times. First application of the reduction formula (n = 3): f(In x)³ dx =+Sdx. Second application of the reduction formula (n = 2): S(In x)³ dx = + Sdx. Third application of the reduction formula: (n = 1) (In)³dx=+S dx. Wrap-up: So completing the final integration above, S(In z)³dx=+C. "¹dx to evaluate f(lnx) ³dx. To achieve this, you will need to apply the reduction formula 3

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