Evaluate ∫ log x dx.

Indefinite Integrals · Question 78 of 90

Question 78

MediumOriginal question

Evaluate $\int \log x\,dx$ .

x \left(\log x + 1\right) + C

Correct answerYour answer

x \log x + C

Correct answerYour answer

- x \left(\log x - 1\right) + C

Correct answerYour answer

x \left(\log x - 1\right) + C

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

By parts, $\int u\,dv = uv - \int v\,du$ , with $u = \log x$ and $dv = 1\,dx$ , so $v = x$ .
$x \log x - \int 1\,dx = x \left(\log x - 1\right) + C$ .

Worth noting

Adding the second integral instead of subtracting it is the slip one of the options is built on.

Ten questions from Indefinite Integrals, at your own pace.

Ten questions from Indefinite Integrals, at your own pace.