Choose the correct answer of the given question
`int 1/(1+tanx) dx = ___ +c`

Options

A. log |sexc-tanx|
B. ` 2sec^2 x/2`
C. log|x+sinx|
D. 1/2[x+log|sinx+cosx|]

Answer

To solve the integral \( \int \frac{1}{1 + \tan x} \, dx \), we will use substitution and partial fractions. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( z = \tan x \). Then, the differential \( dz \) is given by:
\[ dz = \sec^2 x \, dx \]
Since \( \sec^2 x = 1 + \tan^2 x = 1 + z^2 \), we can express \( dx \) as:
\[ dx = \frac{dz}{1 + z^2} \]

### Step 2: Rewrite the Integral
Substituting \( z \) into the integral, we have:
\[ \int \frac{1}{1 + \tan x} \, dx = \int \frac{1}{1 + z} \cdot \frac{dz}{1 + z^2} \]
This simplifies to:
\[ \int \frac{dz}{(1 + z)(1 + z^2)} \]

### Step 3: Partial Fraction Decomposition
We can decompose the integrand using partial fractions:
\[ \frac{1}{(1 + z)(1 + z^2)} = \frac{A}{1 + z} + \frac{Bz + C}{1 + z^2} \]
Multiplying through by the denominator \( (1 + z)(1 + z^2) \) gives:
\[ 1 = A(1 + z^2) + (Bz + C)(1 + z) \]
Expanding this, we get:
\[ 1 = A + Az^2 + Bz + Bz^2 + C + Cz \]
Combining like terms:
\[ 1 = (A + B)z^2 + (B + C)z + (A + C) \]
Setting coefficients equal, we have:
1. \( A + B = 0 \)
2. \( B + C = 0 \)
3. \( A + C = 1 \)

### Step 4: Solve for A, B, and C
From \( A + B = 0 \), we have \( B = -A \).
From \( B + C = 0 \), substituting \( B \) gives \( -A + C = 0 \) or \( C = A \).
Substituting \( C \) into \( A + C = 1 \) gives:
\[ A + A = 1 \implies 2A = 1 \implies A = \frac{1}{2} \]
Thus, \( B = -\frac{1}{2} \) and \( C = \frac{1}{2} \).

### Step 5: Rewrite the Integral
Now, substituting back into the integral:
\[ \int \left( \frac{1/2}{1 + z} + \frac{-1/2 z + 1/2}{1 + z^2} \right) dz \]
This becomes:
\[ \frac{1}{2} \int \frac{1}{1 + z} \, dz - \frac{1}{2} \int \frac{z}{1 + z^2} \, dz + \frac{1}{2} \int \frac{1}{1 + z^2} \, dz \]

### Step 6: Solve Each Integral
1. \( \int \frac{1}{1 + z} \, dz = \ln |1 + z| \)
2. For \( \int \frac{z}{1 + z^2} \, dz \), use substitution \( u = 1 + z^2 \) which gives \( \frac{1}{2} \ln |1 + z^2| \).
3. \( \int \frac{1}{1 + z^2} \, dz = \tan^{-1}(z) \)

Putting it all together:
\[ \frac{1}{2} \ln |1 + z| - \frac{1}{4} \ln |1 + z^2| + \frac{1}{2} \tan^{-1}(z) + C \]

### Step 7: Substitute Back
Substituting \( z = \tan x \):
\[ = \frac{1}{2} \ln |1 + \tan x| - \frac{1}{4} \ln |1 + \tan^2 x| + \frac{1}{2} \tan^{-1}(\tan x) + C \]
Since \( \tan^{-1}(\tan x) = x \):
\[ = \frac{1}{2} \ln |1 + \tan x| - \frac{1}{4} \ln |\sec^2 x| + \frac{1}{2} x + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{1}{1 + \tan x} \, dx = \frac{1}{2} \ln |1 + \tan x| - \frac{1}{4} \ln |\sec^2 x| + \frac{1}{2} x + C \]