Answer
$$\int_{e}^{y} \frac{y \ln x}{x} d x
=\frac{y}{2}\left[(\ln y)^{2}-y^{2}\right],(y\in (0,\infty))$$
Work Step by Step
Given $$\int_{e}^{y} \frac{y \ln x}{x} d x$$
So,
\begin{align}
\int_{e}^{y} \frac{y \ln x}{x} d x &=\left[\frac{1}{2} y \ln ^{2} x\right]_{e^{y}}^{y}\\
&=\frac{1}{2} y\left[\ln ^{2} y-\ln ^{2} e^{y}\right]\\
&=\frac{y}{2}\left[(\ln y)^{2}-y^{2}\right],(y \in(0,\infty))
\end{align}
