Question

Let $I_1=\underset{0}{\overset{1}{\int }}\frac{e^{x}dx}{1+x}$ and $I_2=\underset{0}{\overset{1}{\int }}\frac{x^{2}dx}{e^{x^3}(2-x^3)}$, then $\frac{I_1}{I_2}$ is equal

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

Answer 1

The correct option is C $3e$

$I_1=\underset{0}{\overset{1}{\int }}\frac{e^{x}dx}{1+x},I_2=\underset{0}{\overset{1}{\int }}\frac{x^{2}dx}{e^{x^3}(2-x^3)}$
In $I_2$, put $1-x^3=t$
$\Rightarrow I_2=\frac{1}{3}\underset{1}{\overset{0}{\int }}\frac{-dt}{e^{1-t}(1+t)}$
$=\frac{1}{3e}\underset{0}{\overset{1}{\int }}\frac{e^{t}dt}{1+t}=\frac{1}{3e}{I}_1$
$\Rightarrow \frac{I_1}{I_2}=3e$