Let `I_(1) = int_(0)^(pi//4)1/((1+tanx)^(2))dx`, `I_(2) = int_(0)^(1)(dx)/((1+x)^(2)(1+x^(2)))` Then find the value of `(I_(1))/(I_(2))`.

Answer

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 \( I_1 \)

The integral \( I_1 \) is given by:

\[ I_1 = \int_{0}^{\frac{\pi}{4}} \frac{1}{(1 + \tan x)^2} \, dx \]

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

The integral \( I_2 \) is given by:

\[ I_2 = \int_{0}^{1} \frac{1}{(1 + x)^2 (1 + x^2)} \, dx \]

To simplify \( I_2 \), we can use the substitution \( x = \tan t \). Then, \( dx = \sec^2 t \, dt \) and the limits change from \( x = 0 \) to \( t = 0 \) and from \( x = 1 \) to \( t = \frac{\pi}{4} \).

Substituting into \( I_2 \):

\[ I_2 = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2 t}{(1 + \tan t)^2 (1 + \tan^2 t)} \, dt \]

Using the identity \( 1 + \tan^2 t = \sec^2 t \), we can rewrite \( I_2 \):

\[ I_2 = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2 t}{(1 + \tan t)^2 \sec^2 t} \, dt = \int_{0}^{\frac{\pi}{4}} \frac{1}{(1 + \tan t)^2} \, dt \]

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

Now we observe that:

\[ I_1 = \int_{0}^{\frac{\pi}{4}} \frac{1}{(1 + \tan x)^2} \, dx \]
\[ I_2 = \int_{0}^{\frac{\pi}{4}} \frac{1}{(1 + \tan t)^2} \, dt \]

Since both integrals are equal, we have:

\[ I_1 = I_2 \]

### Step 4: Calculate \( \frac{I_1}{I_2} \)

Thus, we can conclude:

\[ \frac{I_1}{I_2} = \frac{I_2}{I_2} = 1 \]

### Final Answer

The value of \( \frac{I_1}{I_2} \) is:

\[ \boxed{1} \]