Question

Let $I_1={\int }_0^1\frac{e^{x}dx}{1+x}$ and $I_2={\int }_0^1\frac{x^{2}dx}{e^{x^3}(2-x^3)}$, then $\frac{I_1}{I_2}$ is

• A. $3/e$
• B. $e/3$
• C. $3e$
• D. $1/3e$

Answer 1

Answer: (C)

$3e$

Given : $I_1={\int }_0^1\frac{e^{x}dx}{(1+x)}{I}_2={\int }_0^1\frac{x^{2}dx}{e^{x^3}(2-x^3)}$

$I_1={\int }_0^1\frac{e^{x}dx}{(1+x)}{I}_2={\int }_0^1\frac{x^{2}dx}{e^{x^3}(2-x^3)}$

Let : $1-x^3=tx\to 0t\to 1$

$-3x^2=dtx\to 1t\to 0I_2=-{\int }_0^1\frac{1}{3}\frac{e^{t-1}}{(1+t)}dt=\frac{1}{3e}{\int }_0^1\frac{e^t}{(1+t)}dt\frac{I_1}{I_2}=\frac{{\int }_0^1\frac{e^{x}dx}{(1+x)}}{\frac{1}{3e}{\int }_0^1\frac{e^t}{(1+t)}dt}=3e$

Hence the correct answer is $3e$