Key Concepts and Formulas

• Trigonometric Identities: $\tan x = \frac{\sin x}{\cos x}$
• Integral of the form $\int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + C$
• Linear combination technique to solve integrals of the form $\int \frac{a \cos x + b \sin x}{c \cos x + d \sin x} dx$

Step-by-Step Solution

Step 1: Rewrite the integrand in terms of sine and cosine.

We are given the integral $\int \frac{2-\tan x}{3+\tan x} dx$. We rewrite the integrand using the identity $\tan x = \frac{\sin x}{\cos x}$:

$\int \frac{2-\tan x}{3+\tan x} dx = \int \frac{2 - \frac{\sin x}{\cos x}}{3 + \frac{\sin x}{\cos x}} dx = \int \frac{2\cos x - \sin x}{3\cos x + \sin x} dx$

Step 2: Express the numerator as a linear combination of the denominator and its derivative.

Let $2\cos x - \sin x = a(3\cos x + \sin x) + b(-3\sin x + \cos x)$, where $a$ and $b$ are constants. Here, $3 \cos x + \sin x$ is the denominator and $-3 \sin x + \cos x$ is its derivative.

Expanding the right side gives: $2\cos x - \sin x = (3a+b)\cos x + (a-3b)\sin x$.

Comparing the coefficients of $\cos x$ and $\sin x$ on both sides, we obtain the following system of equations:

$3a + b = 2 \quad \text{.... (1)}$ $a - 3b = -1 \quad \text{.... (2)}$

Step 3: Solve the system of equations for a and b.

Multiply equation (1) by 3: $9a + 3b = 6 \quad \text{.... (3)}$ Add equation (2) and (3): $10a = 5$ $a = \frac{1}{2}$

Substitute $a = \frac{1}{2}$ into equation (1): $3\left(\frac{1}{2}\right) + b = 2$ $\frac{3}{2} + b = 2$ $b = 2 - \frac{3}{2} = \frac{1}{2}$

Thus, $a = \frac{1}{2}$ and $b = \frac{1}{2}$.

Step 4: Substitute the values of a and b back into the integral.

Now we can rewrite the integral as:

$\int \frac{2\cos x - \sin x}{3\cos x + \sin x} dx = \int \frac{\frac{1}{2}(3\cos x + \sin x) + \frac{1}{2}(-3\sin x + \cos x)}{3\cos x + \sin x} dx$ $= \frac{1}{2} \int \frac{3\cos x + \sin x}{3\cos x + \sin x} dx + \frac{1}{2} \int \frac{-3\sin x + \cos x}{3\cos x + \sin x} dx$ $= \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{-3\sin x + \cos x}{3\cos x + \sin x} dx$

Step 5: Evaluate the integrals.

The first integral is straightforward: $\int 1 dx = x + C_1$.

For the second integral, notice that the numerator is the derivative of the denominator. Therefore, $\int \frac{-3\sin x + \cos x}{3\cos x + \sin x} dx = \ln |3\cos x + \sin x| + C_2$

So, the integral becomes:

$\int \frac{2-\tan x}{3+\tan x} dx = \frac{1}{2}x + \frac{1}{2}\ln |3\cos x + \sin x| + C$ where $C = \frac{1}{2}C_1 + \frac{1}{2}C_2$.

Step 6: Compare with the given form and find the values of α, β, and γ.

We are given that $\int \frac{2-\tan x}{3+\tan x} dx = \frac{1}{2}\left(\alpha x + \log_e |\beta \sin x + \gamma \cos x|\right) + C$. Comparing this with our result, $\frac{1}{2}x + \frac{1}{2}\ln |3\cos x + \sin x| + C$, we have:

$\alpha = 1$ $\beta = 1$ $\gamma = 3$

Step 7: Calculate α + γ/β.

$\alpha + \frac{\gamma}{\beta} = 1 + \frac{3}{1} = 1 + 3 = 4$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when differentiating and substituting. A small sign error can propagate through the entire solution.
• Choosing the right linear combination: The key is to express the numerator as a combination of the denominator and its derivative. This makes the integration straightforward.
• Don't forget the constant of integration: Always include the constant of integration, C, in indefinite integrals.

Summary

We solved the indefinite integral $\int \frac{2-\tan x}{3+\tan x} dx$ by first converting the integrand to sine and cosine. Then, we expressed the numerator as a linear combination of the denominator and its derivative, allowing us to split the integral into simpler parts. After evaluating the integrals and comparing the result with the given form, we found the values of $\alpha$, $\beta$, and $\gamma$. Finally, we calculated $\alpha + \frac{\gamma}{\beta}$, which equals 4.

Final Answer The final answer is \boxed{4}, which corresponds to option (D).