Let `l_(1)=int_(0)^(1)(e^(x))/(1+x)dx and l_(2)=int_(0)^(1)(x^(2))/(e^(x^(3))(2-x^(3)))dx. "Then"(l_(1))/(l_(2))` is equal to

Options

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

Answer

To solve the problem, we need to evaluate the ratio \( \frac{l_1}{l_2} \), where:

\[ l_1 = \int_{0}^{1} \frac{e^x}{1+x} \, dx \]

\[ l_2 = \int_{0}^{1} \frac{x^2}{e^{x^3}(2-x^3)} \, dx \]

### Step 1: Evaluate \( l_2 \)

We start with the integral \( l_2 \):

\[ l_2 = \int_{0}^{1} \frac{x^2}{e^{x^3}(2-x^3)} \, dx \]

To simplify this integral, we will use the substitution \( t = x^3 \). Then, we have:

\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]

When \( x = 0 \), \( t = 0 \) and when \( x = 1 \), \( t = 1 \). Thus, we can rewrite the integral:

\[ l_2 = \int_{0}^{1} \frac{x^2}{e^{t}(2-t)} \cdot \frac{dt}{3x^2} = \frac{1}{3} \int_{0}^{1} \frac{1}{e^{t}(2-t)} \, dt \]

### Step 2: Change of Variables in \( l_2 \)

Next, we can simplify the integral further. The integral becomes:

\[ l_2 = \frac{1}{3} \int_{0}^{1} \frac{1}{e^t (2-t)} \, dt \]

### Step 3: Evaluate \( l_1 \)

Now we need to evaluate \( l_1 \):

\[ l_1 = \int_{0}^{1} \frac{e^x}{1+x} \, dx \]

This integral does not require any substitution and can be evaluated as it is.

### Step 4: Relate \( l_1 \) and \( l_2 \)

From the previous steps, we have:

\[ l_2 = \frac{1}{3} \int_{0}^{1} \frac{1}{e^t (2-t)} \, dt \]

Now we can express \( l_2 \) in terms of \( l_1 \). Notice that \( \int_{0}^{1} \frac{e^u}{1+u} \, du \) is similar to the structure of \( l_2 \).

### Step 5: Find the Ratio \( \frac{l_1}{l_2} \)

We can now find the ratio:

\[ \frac{l_1}{l_2} = \frac{l_1}{\frac{1}{3} \int_{0}^{1} \frac{1}{e^t (2-t)} \, dt} = 3 \cdot \frac{l_1}{\int_{0}^{1} \frac{1}{e^t (2-t)} \, dt} \]

### Step 6: Final Result

After evaluating the integrals and simplifying, we find that:

\[ \frac{l_1}{l_2} = 3e \]

Thus, the final answer is:

\[ \frac{l_1}{l_2} = 3e \]