Let {I_1} = intlimits_0^1 {frac{{{e^x}}}{{1 + | vedclass

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(C) Given $I_2 = \int_0^1 \frac{x^2}{e^{x^3}(2-x^3)} dx$. Let $t = x^3$,then $dt = 3x^2 dx$,so $x^2 dx = \frac{1}{3} dt$. When $x=0, t=0$ and when $x=

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$3e$

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(C) Given $I_2 = \int_0^1 \frac{x^2}{e^{x^3}(2-x^3)} dx$. Let $t = x^3$,then $dt = 3x^2 dx$,so $x^2 dx = \frac{1}{3} dt$. When $x=0, t=0$ and when $x=1, t=1$. Thus,$I_2 = \frac{1}{3} \int_0^1 \frac{dt}{e^t(2-t)} = \frac{1}{3} \int_0^1 \frac{e^{-t}}{2-t} dt$. Let $u = 1-t$,then $du = -dt$. When $t=0, u=1$ and when $t=1, u=0$. $I_2 = \frac{1}{3} \int_1^0 \frac{e^{-(1-u)}}{2-(1-u)} (-du) = \frac{1}{3} \int_0^1 \frac{e^{u-1}}{1+u} du = \frac{1}{3e} \int_0^1 \frac{e^u}{1+u} du$. Since $I_1 = \int_0^1 \frac{e^x}{1+x} dx$,we have $I_2 = \frac{1}{3e} I_1$. Therefore,$\frac{I_1}{I_2} = 3e$.

Let ${I_1} = \int\limits_0^1 {\frac{{{e^x}}}{{1 + x}}} \,dx$ and ${I_2} = \int\limits_0^1 {\frac{{{x^2}}}{{{e^{{x^3}}}\left( {2 - {x^3}} \right)}}} \,dx$,then the value of $\frac{{{I_1}}}{{{I_2}}}$ is equal to

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