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 equal to?

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

Answer 1

The correct option is C $3e$

In $I_2={\int }_0^1\frac{x^{2}dx}{e^{x^3}(2-x^3)}$
Substitute $1-x^3=t\Rightarrow -3x^{2}dx=dt$
We get $I_2=-\frac{1}{3}{\int }_1^0\frac{dt}{e^{1-t}(1+t)}=\frac{1}{3e}{\int }_0^1\frac{e^{t}dt}{1+t}=\frac{1}{3e}{\int }_0^1\frac{e^{x}dx}{1+x}$
As $I_1={\int }_0^1\frac{e^{x}dx}{1+x}$
Therefore $\frac{I_1}{I_2}=3e$