`int_(0)^(pi//4) (sec^(2)x)/((1+tan x)(2+tan x))dx` is equal to

Options

A. `log_(e)((2)/(3))`
B. `log_(e)3`
C. `(1)/(3)log_(e)((4)/(3))`
D. `log_(e)((4)/(3))`

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2 x}{(1 + \tan x)(2 + \tan x)} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = 1 + \tan x \). Then, we differentiate:
\[ dt = \sec^2 x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\sec^2 x} \]
When \( x = 0 \), \( t = 1 + \tan(0) = 1 \). When \( x = \frac{\pi}{4} \), \( t = 1 + \tan\left(\frac{\pi}{4}\right) = 2 \).

### Step 2: Change the limits and rewrite the integral
Now, we can rewrite the integral in terms of \( t \):
\[ I = \int_{1}^{2} \frac{1}{t(1 + t)} \, dt \]

### Step 3: Partial Fraction Decomposition
We can decompose \( \frac{1}{t(1 + t)} \) into partial fractions:
\[ \frac{1}{t(1 + t)} = \frac{A}{t} + \frac{B}{1 + t} \]
Multiplying through by \( t(1 + t) \):
\[ 1 = A(1 + t) + Bt \]
Setting \( t = 0 \):
\[ 1 = A \quad \Rightarrow \quad A = 1 \]
Setting \( t = -1 \):
\[ 1 = B(-1) \quad \Rightarrow \quad B = -1 \]
Thus, we have:
\[ \frac{1}{t(1 + t)} = \frac{1}{t} - \frac{1}{1 + t} \]

### Step 4: Integrate
Now we can rewrite the integral:
\[ I = \int_{1}^{2} \left( \frac{1}{t} - \frac{1}{1 + t} \right) dt \]
This can be split into two integrals:
\[ I = \int_{1}^{2} \frac{1}{t} \, dt - \int_{1}^{2} \frac{1}{1 + t} \, dt \]
Calculating these integrals:
\[ \int \frac{1}{t} \, dt = \ln |t| \quad \text{and} \quad \int \frac{1}{1 + t} \, dt = \ln |1 + t| \]
Thus,
\[ I = \left[ \ln t \right]_{1}^{2} - \left[ \ln(1 + t) \right]_{1}^{2} \]
Calculating the limits:
\[ I = \left( \ln 2 - \ln 1 \right) - \left( \ln(3) - \ln(2) \right) \]
Since \( \ln 1 = 0 \):
\[ I = \ln 2 - (\ln 3 - \ln 2) = \ln 2 - \ln 3 + \ln 2 = 2\ln 2 - \ln 3 \]

### Final Result
Thus, the value of the integral is:
\[ I = \ln\left(\frac{4}{3}\right) \]