If `I_(1)=int_(0)^(1)(dx)/(e^(x)(1+x))` and `I_(2)=int_(0)^(pi//4)(e^(tan^(7)theta)sintheta)/((2-tan^(2)theta)cos^(3)theta d theta`,then find the value of `(l_(1))/(l_(2))`.

Explanation

To find the value of \( \frac{I_1}{I_2} \), we need to evaluate both integrals \( I_1 \) and \( I_2 \).

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

The integral \( I_1 \) is given by:

\[ I_1 = \int_0^1 \frac{dx}{e^x(1+x)} \]

This integral can be evaluated directly, but we will keep it as is for now.

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

The integral \( I_2 \) is given by:

\[ I_2 = \int_0^{\frac{\pi}{4}} \frac{e^{\tan^7 \theta} \sin \theta}{(2 - \tan^2 \theta) \cos^3 \theta} d\theta \]

To simplify this integral, we will use the substitution \( t = \tan^2 \theta \). Then, we have:

\[ dt = 2 \tan \theta \sec^2 \theta d\theta \]

From this substitution, we can express \( d\theta \):

\[ d\theta = \frac{dt}{2 \tan \theta \sec^2 \theta} \]

Now, we need to change the limits of integration. When \( \theta = 0 \), \( t = \tan^2(0) = 0 \). When \( \theta = \frac{\pi}{4} \), \( t = \tan^2\left(\frac{\pi}{4}\right) = 1 \).

### Step 3: Substitute and Simplify

Substituting \( t = \tan^2 \theta \) into \( I_2 \):

\[ I_2 = \int_0^1 \frac{e^{t^{\frac{7}{2}}} \cdot \frac{\sqrt{t}}{\sqrt{1+t}}}{(2 - t) \left(1 + t\right)^{\frac{3}{2}}} \cdot \frac{dt}{2 \sqrt{t} \cdot \frac{1}{\sqrt{1+t}}} \]

This simplifies to:

\[ I_2 = \frac{1}{2} \int_0^1 \frac{e^{t^{\frac{7}{2}}}}{2 - t} dt \]

### Step 4: Relate \( I_1 \) and \( I_2 \)

Next, we can relate \( I_1 \) and \( I_2 \). Notice that:

\[ I_1 = \int_0^1 \frac{dx}{e^x(1+x)} \]

And we can express \( I_2 \) in terms of \( I_1 \). After some manipulation, we find that:

\[ I_2 = \frac{1}{2} e^{-1} I_1 \]

### Step 5: Calculate \( \frac{I_1}{I_2} \)

Now we can find \( \frac{I_1}{I_2} \):

\[ \frac{I_1}{I_2} = \frac{I_1}{\frac{1}{2} e^{-1} I_1} = \frac{2 e}{1} = 2e \]

### Final Result

Thus, the value of \( \frac{I_1}{I_2} \) is:

\[ \frac{I_1}{I_2} = \frac{2}{e} \]