Let `int frac{2 – tan x}{3 + tan x} dx = 1/2 (alpha x + log_e |beta sin x + gamma cos x|) + C`,
where `C` is the constant of integration. Then `alpha + (gamma)/(beta)` is equal to is equal to

Options

A. 1
B. 4
C. 3
D. 7

Answer

To solve the given integral problem step by step, we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{2 - \tan x}{3 + \tan x} \, dx \]
We can express \(\tan x\) in terms of \(\sin x\) and \(\cos x\):
\[ \tan x = \frac{\sin x}{\cos x} \]
Thus, we rewrite the integral as:
\[ \int \frac{2 - \frac{\sin x}{\cos x}}{3 + \frac{\sin x}{\cos x}} \, dx = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} \, dx \]

### Step 2: Use Partial Fraction Decomposition
We can express the numerator \(2 \cos x - \sin x\) as a linear combination of the derivative of the denominator:
\[ 2 \cos x - \sin x = a(3 \cos x + \sin x) + b(3 \cos x + \sin x) \]
We need to find constants \(a\) and \(b\) such that:
\[ 2 \cos x - \sin x = a(3 \cos x - \sin x) + b(3 \cos x + \sin x) \]

### Step 3: Set Up the System of Equations
By comparing coefficients of \(\cos x\) and \(\sin x\), we get:
1. For \(\cos x\): \(2 = 3a + 3b\)
2. For \(\sin x\): \(-1 = -a + b\)

### Step 4: Solve the System of Equations
From the second equation, we can express \(b\) in terms of \(a\):
\[ b = a - 1 \]
Substituting \(b\) into the first equation:
\[ 2 = 3a + 3(a - 1) \implies 2 = 6a - 3 \implies 6a = 5 \implies a = \frac{5}{6} \]
Now substituting \(a\) back to find \(b\):
\[ b = \frac{5}{6} - 1 = -\frac{1}{6} \]

### Step 5: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \left( \frac{5/6}{3 + \tan x} - \frac{1/6}{3 + \tan x} \right) \, dx \]
This can be separated into two integrals:
\[ \frac{5}{6} \int \frac{1}{3 + \tan x} \, dx - \frac{1}{6} \int \frac{1}{3 + \tan x} \, dx \]

### Step 6: Integrate
The integral of \(\frac{1}{3 + \tan x}\) can be solved using the substitution \(u = 3 + \tan x\):
\[ du = \sec^2 x \, dx \]
Thus, the integral becomes:
\[ \int \frac{1}{u} \, du = \log |u| + C = \log |3 + \tan x| + C \]

### Step 7: Combine Results
Combining the results, we have:
\[ \int \frac{2 - \tan x}{3 + \tan x} \, dx = \frac{1}{2} \left( \alpha x + \log |3 \cos x + \sin x| \right) + C \]
where \(\alpha = 1\), \(\beta = 1\), and \(\gamma = 3\).

### Step 8: Calculate \(\alpha + \frac{\gamma}{\beta}\)
Now we need to compute:
\[ \alpha + \frac{\gamma}{\beta} = 1 + \frac{3}{1} = 1 + 3 = 4 \]

### Final Answer
Thus, the value of \(\alpha + \frac{\gamma}{\beta}\) is:
\[ \boxed{4} \]