If `I_1=int_0^1(e^x)/(1+x) dx`aand `I_2=int_0^1 x^2/(e^(x^3)(2-x^3)) dx` then `I_1/I_2` is

Options

A. `3//e`
B. `e//3`
C. `3e`
D. `1//3e`

Explanation

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and then find the ratio \( \frac{I_1}{I_2} \).

### Step 1: Evaluate \( I_1 \)

Given:
\[ I_1 = \int_0^1 \frac{e^x}{1+x} \, dx \]

This integral does not have a straightforward elementary solution, so we will keep it as is for now.

### Step 2: Evaluate \( I_2 \)

Given:
\[ I_2 = \int_0^1 \frac{x^2}{e^{x^3}(2-x^3)} \, dx \]

To simplify this, we use the substitution \( t = 1 - x^3 \). Then, we have:
\[ x^3 = 1 - t \quad \Rightarrow \quad dx = -\frac{dt}{3x^2} \]
From the substitution, when \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 0 \).

Now, we need to express \( x^2 \) in terms of \( t \):
\[ x^2 = (1 - t)^{2/3} \]

Thus, we can rewrite \( I_2 \):
\[ I_2 = \int_1^0 \frac{(1-t)^{2/3}}{e^{1-t}(2 - (1-t))} \left(-\frac{dt}{3(1-t)^{2/3}}\right) \]
This simplifies to:
\[ I_2 = \frac{1}{3} \int_0^1 \frac{1}{e^{1-t}(1+t)} \, dt \]

### Step 3: Relate \( I_1 \) and \( I_2 \)

Notice that:
\[ I_1 = \int_0^1 \frac{e^x}{1+x} \, dx \]
and we have transformed \( I_2 \) into:
\[ I_2 = \frac{1}{3} \int_0^1 \frac{e^t}{t+1} \, dt \]

### Step 4: Find the ratio \( \frac{I_1}{I_2} \)

Now we can compute the ratio:
\[ \frac{I_1}{I_2} = \frac{\int_0^1 \frac{e^x}{1+x} \, dx}{\frac{1}{3} \int_0^1 \frac{e^t}{t+1} \, dt} \]
This simplifies to:
\[ \frac{I_1}{I_2} = 3 \cdot \frac{\int_0^1 \frac{e^x}{1+x} \, dx}{\int_0^1 \frac{e^t}{t+1} \, dt} \]

Since the integrals are equal, we find:
\[ \frac{I_1}{I_2} = 3 \]

### Conclusion

Thus, the final result is:
\[ \frac{I_1}{I_2} = 3 \]