Let I1=∫01ex1+xdx and I2=∫01x2ex3(2−x3)dx then I1eI2 is

I2=13∫01 3x2ex3(2−x3)dx Put x3=tI2=13∫01 dtet(2−t)=13e∫01 et(1+t)dt=13e∫01 ex1+xdx=13eI1

Let I1=∫01ex1+xdx and I2=∫01x2ex3(2−x3)dx then I1eI2 is

I2=13∫01 3x2ex3(2−x3)dx Put x3=tI2=13∫01 dtet(2−t)=13e∫01 et(1+t)dt=13e∫01 ex1+xdx=13eI1

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Let $I_{1} = \int_{0}^{1} \frac{e^{x}}{1 + x} d x$ and $I_{2} = \int_{0}^{1} \frac{x^{2}}{e^{x^{3}} (2 - x^{3})} d x$ then $\frac{I_{1}}{e I_{2}}$ is

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Detailed Solution

$I_{2} = \frac{1}{3} \int_{0}^{1} \frac{3 x^{2}}{e^{x^{3}} (2 - x^{3})} d x$ Put $x^{3} = t$
$I_{2} = \frac{1}{3} \int_{0}^{1} \frac{d t}{e^{t} (2 - t)} = \frac{1}{3 e} \int_{0}^{1} \frac{e^{t}}{(1 + t)} d t = \frac{1}{3 e} \int_{0}^{1} \frac{e^{x}}{1 + x} d x = \frac{1}{3 e} I_{1}$