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:

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

To evaluate this integral, we can use the substitution \( t = \tan x \). Therefore, \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{1 + t^2} \). The limits change as follows:
- When \( x = 0 \), \( t = \tan(0) = 0 \).
- When \( x = \frac{\pi}{4} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \).

Now, substituting these into the integral:

\[ I_1 = \int_{0}^{1} \frac{1}{(1 + t)^2} \cdot \frac{dt}{1 + t^2} \]

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

The integral \( I_2 \) is given by:

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

This integral can be evaluated directly or using a similar substitution as above.

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

Notice that both \( I_1 \) and \( I_2 \) have the same form after substitution:

\[ I_1 = \int_{0}^{1} \frac{1}{(1 + t)^2 (1 + t^2)} \, dt \]

Thus, we find that:

\[ I_1 = I_2 \]

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

Since \( I_1 = I_2 \), we can conclude:

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

### Final Answer

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

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