The value of ∫_1^e((log_e x)^2)/x dx is

The correct answer is Option (2) → \(\displaystyle\frac{1}{3}\)

Let \(t = \log_e x\). Then

    \( \displaystyle dt = \frac{1}{x}\,dx \)

The limits change with the substitution:

    \( x = 1 \Rightarrow t = 0 \)
    \( x = e \Rightarrow t = 1 \)

So

    \( \displaystyle \int_{1}^{e}\frac{(\log_e x)^{2}}{x}\,dx = \int_{0}^{1}t^{2}\,dt \)

    \( \displaystyle = \Big[\frac{t^{3}}{3}\Big]_{0}^{1} = \frac{1}{3} \)