For \(x \gt 1\), \(\int\left(\log(\log x) + \displaystyle\frac{1}{\log x}\right)dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(x\log(\log x) + C\)
With \(\log(\log x)\) as the first function and 1 as the second, integration by parts gives
\( \displaystyle \int \log(\log x) \cdot 1\,dx = x\log(\log x) - \int x \cdot \frac{1}{\log x} \cdot \frac{1}{x}\,dx \)
\( \displaystyle = x\log(\log x) - \int \frac{dx}{\log x} \)
Adding \(\int \frac{dx}{\log x}\), the two integrals cancel:
\( \displaystyle \int\left(\log(\log x) + \frac{1}{\log x}\right)dx = x\log(\log x) + C \)
Check: \(\frac{d}{dx}\left[x\log(\log x)\right] = \log(\log x) + x \cdot \frac{1}{x\log x} = \log(\log x) + \frac{1}{\log x}\).