∫_1^exlog_e x dx is equal to:

The correct answer is Option (2) → \( \displaystyle\frac{1}{4}(e^2+1) \)

Integrating by parts with \( u=\log_e x \) and \( dv=x\,dx \), so that \( du=\frac{dx}{x} \) and \( v=\frac{x^2}{2} \):

    \( \displaystyle \int x\log_e x\,dx=\frac{x^2}{2}\log_e x-\int\frac{x^2}{2}\cdot\frac{1}{x}\,dx \)

    \( \displaystyle =\frac{x^2}{2}\log_e x-\frac{x^2}{4} \)

Putting in the limits, with \( \log_e e=1 \) and \( \log_e 1=0 \):

    \( \displaystyle \int_{1}^{e}x\log_e x\,dx=\left(\frac{e^2}{2}-\frac{e^2}{4}\right)-\left(0-\frac{1}{4}\right) \)

    \( \displaystyle =\frac{e^2}{4}+\frac{1}{4} \)

    \( \displaystyle =\frac{1}{4}(e^2+1) \)

Tip: The lower limit is where sign slips happen: at x = 1 the antiderivative is \( -\frac{1}{4} \), and subtracting it adds \( \frac{1}{4} \). Dropping that sign gives \( \frac{1}{4}(e^2-1) \), Option (3).