To evaluate $\int x^2\log x\,dx$ by parts, which should be taken as $u$ , the function to differentiate?

\log x

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x^2

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x^2\log x

Correct answerYour answerEither choice works equally wellCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

We cannot integrate $\log x$ easily, but we can differentiate it to $\frac{1}{x}$ , which cancels a power of $x$ .
So $u = \log x$ and $dv = x^2\,dx$ . The order Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential (ILATE) puts the logarithm before the algebraic factor.

Worth noting

With $u = x^2$ , $dv = \log x\,dx$ has to be integrated first, which is itself a by-parts problem.

Ten questions from Indefinite Integrals, at your own pace.

To evaluate ∫x2log⁡x dx\int x^2\log x\,dx∫x2logxdx by parts, which should be taken as uuu, the function to differentiate?