\(\displaystyle\int \frac{(\log x)^2}{x}\,dx,\ x \gt 0\), is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{(\log x)^3}{3} + C\)
Since \(\frac{d}{dx}\log x = \frac{1}{x}\), the substitution \(t = \log x\) fits.
\( \displaystyle dt = \frac{1}{x}\,dx \)
Then
\( \displaystyle \int \frac{(\log x)^2}{x}\,dx = \int t^2\,dt \)
\( \displaystyle = \frac{t^3}{3} + C \)
\( \displaystyle = \frac{(\log x)^3}{3} + C \)