Use integration by parts to establish the reduction formula: [(In x)" daz = 2(lnx)" - n [ (in z)-¹da: To begin the integration by parts, let u = and du = =dz and v=[ leads immediately to the above reduction formula. Evaluating du = Now apply the reduction formula to the problem: (Inz)ºdz. Getting the final result would be tedious, requiring 9 applications of the reduction formula. So for simplicity just give the result of applying the reduction formula one time. (In a)'da is simplified to 0+/Odz dx +C.

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