Don't be intimidated!

Substituting $\displaystyle e^{x^2}-1=t$ the integral changes to

$\int_0^1\frac{\ln\left(1+t\right)}{t}dt$

$=\int_0^1\sum_{n=1}^{\infty}\frac{\left(-t\right)^{n-1}}{n}dt$

$=\sum_{n=1}^{\infty}\left(\int_0^1\frac{\left(-t\right)^{n-1}}{n}dt\right)$

$=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n^2}$

$=\left(\sum_{n=1}^{\infty}\frac{1}{n^2}\right)-\left(\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}\right)$

$=\frac{\pi^2}{12}$

Thus, making the answer $\displaystyle 1\cdot2\cdot12=24$

Note

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ follows from the Riemann Zeta Function