Question

Let $I_1={\int }_0^1\frac{e^x}{1+x}dx$ and $I_2={\int }_0^1\frac{x^{2}dx}{e^{x^3}(2-x^3)}$. Value of $\frac{I_1}{I_2}$ is equal to

• A. $\frac{3}{e}$
• B. $\frac{e}{3}$
• C. $3e$
• D. $\frac{1}{3e}$

Answer 1

Answer: (C)

$3e$

$I_2=\frac{1}{3}{\int }_0^1\frac{3x^{2}dx}{e^{x^3}(2-x^3)}$, substituting $t=x^3$
$\Rightarrow I_2=\frac{1}{3}{\int }_0^1\frac{dt}{e^t(2-t)}\Rightarrow$ Substituting $t=1-z$
then $I_2=\frac{1}{3}{\int }_0^1\frac{dz}{e^{1-z}(1+z)}=\frac{1}{3e}{\int }_0^1\frac{e^{z}dz}{(z+1)}$
$\Rightarrow \frac{I_1}{I_2}=3e$