Evaluate ∫ (log x)/x^2 dx.

Indefinite Integrals · Question 90 of 90

Question 90

HardOriginal question

Evaluate $\int \frac{\log x}{x^{2}}\,dx$ .

-\frac{\log x}{x} + \frac{1}{x} + C

Correct answerYour answer

-\frac{\log x}{x} - \frac{1}{x} + C

Correct answerYour answer

\frac{\log x}{x} - \frac{1}{x} + C

Correct answerYour answer

\frac{(\log x)^{2}}{2} + C

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

Take $u = \log x$ , $dv = x^{-2}\,dx$ : $v = -\frac{1}{x}$ .
$-\frac{\log x}{x} + \int \frac{1}{x^{2}}\,dx = -\frac{\log x}{x} - \frac{1}{x} + C$ .

Worth noting

$+\frac{1}{x}$ treats $\int x^{-2}\,dx$ as $+\frac{1}{x}$ ; it is $-\frac{1}{x}$ .

Ten questions from Indefinite Integrals, at your own pace.

Ten questions from Indefinite Integrals, at your own pace.