Consider the following statements, where \(x \gt 1\) and C is a constant of integration:
(A) \(\int\log x\,dx = x\log x-x+C\)
(B) \(\int x\log x\,dx = \displaystyle\frac{x^2}{2}\log x-\frac{x^2}{2}+C\)
(C) \(\int\displaystyle\frac{\log x}{x}\,dx = \frac{(\log x)^2}{2}+C\)
(D) \(\int\displaystyle\frac{dx}{x\log x} = \log\left|\log x\right|+C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
The first two need integration by parts with \(\log x\) as the first function:
\( \displaystyle \int\log x\,dx = x\log x-\int x\cdot\frac{1}{x}\,dx = x\log x-x \)
\( \displaystyle \int x\log x\,dx = \frac{x^2}{2}\log x-\int\frac{x}{2}\,dx = \frac{x^2}{2}\log x-\frac{x^2}{4} \)
The last two follow from the substitution \(t = \log x\), \(dt = \frac{dx}{x}\):
\( \displaystyle \int\frac{\log x}{x}\,dx = \int t\,dt = \frac{(\log x)^2}{2} \)
\( \displaystyle \int\frac{dx}{x\log x} = \int\frac{dt}{t} = \log\left|\log x\right| \)
So (A), (C) and (D) agree with these results, while \(\int x\log x\,dx = \frac{x^2}{2}\log x-\frac{x^2}{4}+C\).
Checking the statements
(A) \(\int\log x\,dx = x\log x-x+C\) — Correct. Its derivative is \(\log x+1-1 = \log x\).
(B) \(\int x\log x\,dx = \frac{x^2}{2}\log x-\frac{x^2}{2}+C\) — Incorrect. The last term is \(-\frac{x^2}{4}\); the given result has derivative \(x\log x-\frac{x}{2}\).
(C) \(\int\frac{\log x}{x}\,dx = \frac{(\log x)^2}{2}+C\) — Correct. Its derivative is \(\log x\cdot\frac{1}{x}\).
(D) \(\int\frac{dx}{x\log x} = \log\left|\log x\right|+C\) — Correct. Its derivative is \(\frac{1}{\log x}\cdot\frac{1}{x}\).