3. [4] Find all $y>1$ satisfying $\int_{1}^{y} x \ln x d x=\frac{1}{4}$.
Answer: $\sqrt{\sqrt{e}}$ Applying integration by parts with $u=\ln x$ and $v=\frac{1}{2} x^{2}$, we get
$$
\int_{1}^{y} x \ln x d x=\left.\frac{1}{2} x^{2} \ln x\right|_{1} ^{y}-\frac{1}{2} \int_{1}^{y} x d x=\frac{1}{2} y^{2} \ln y-\frac{1}{4} y^{2}+\frac{1}{4} \text {. }
$$

So $y^{2} \ln y=\frac{1}{2} y^{2}$. Since $y>1$, we obtain $\ln y=\frac{1}{2}$, and thus $y=\sqrt{e}$.