Let $I_1={\int }_0^1 \frac{e^x}{1+x}dx$ and $I_2={\int }_0^1 \frac{x^2}{e^{x^3}\left(2-x^3\right)}dx\text{. }$Then, $\frac{I_1}{l_2},$ is equal to

We have,

$I_1={\int }_0^1 \frac{e^x}{1+x}dx$ and $I_2={\int }_0^1 \frac{x^2}{e^{x^3}\left(2-x^3\right)}dx$

$\therefore l_2=\frac{1}{3}{\int }_0^1 \frac{1}{e^{x^3}\left(2-x^3\right)}3x^{2}dx$

$\Rightarrow I_2=\frac{1}{3}{\int }^1\frac{1}{e^t(2-t)}dt,$where $t=x^3$

$\Rightarrow I_2=\frac{1}{3}{\int }_0^1 \frac{1}{e^{1-t}(1+t)}dt\left[\because {\int }_0^a f\left(x\right)dx={\int }_0^a f(a-x)\right]$

$\Rightarrow I_2=\frac{1}{3e^{\mathrm{\prime }}}{\int }_0^1 \frac{e^t}{1+t}dt\Rightarrow I_2=\frac{1}{3e}{I}_1\Rightarrow \frac{I_1}{I_2}=3e$