Answer
$$\dfrac{2-\sqrt{e}}{6}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^1_0 \int^{\sqrt{e}}_1 \int^e_1 se^s ln r \frac{(\ln t)^2}{t} \space dt \space dr \space ds= \int^1_0 \int^{\sqrt{e}}_1 (se^s \ln r)[\frac{1}{3}(\ln t)^3]^e_1 \space dr \space ds \\=\int^1_0 \int^{\sqrt{e}}_1 \frac{se^s}{3} \ln (r) \space dr \space ds \\=\dfrac{2-\sqrt{e}}{6} \int^1_0 se^s ds \\=\dfrac{2-\sqrt{e}}{6} [se^s-e^s]^1_0 \\=\dfrac{2-\sqrt{e}}{6}$$
