`int (1 + x sin x + cos x)/(x(1 + cos x)) dx = `

Options

A. `"log" x (1 + cos x)`
B. `"log" (x)/(1 + cos x)`
C. `"log" x "sec"^2 x/2`
D. `log x + 2 log ("sec" x/2)`

Answer

To solve the integral

\[ \int \frac{1 + x \sin x + \cos x}{x(1 + \cos x)} \, dx, \]

we can break it down step by step.

### Step 1: Split the Integral
We can separate the integral into three parts:

\[ \int \frac{1}{x(1 + \cos x)} \, dx + \int \frac{x \sin x}{x(1 + \cos x)} \, dx + \int \frac{\cos x}{x(1 + \cos x)} \, dx. \]

This simplifies to:

\[ \int \frac{1}{x(1 + \cos x)} \, dx + \int \frac{\sin x}{1 + \cos x} \, dx + \int \frac{1}{x} \, dx. \]

### Step 2: Solve the First Integral
The first integral is:

\[ \int \frac{1}{x(1 + \cos x)} \, dx. \]

This integral can be solved using the logarithmic identity:

\[ \int \frac{1}{x} \, dx = \log |x| + C. \]

### Step 3: Solve the Second Integral
The second integral is:

\[ \int \frac{\sin x}{1 + \cos x} \, dx. \]

To solve this, we can use the substitution \( u = 1 + \cos x \), which gives \( du = -\sin x \, dx \). Thus,

\[ \int \frac{\sin x}{1 + \cos x} \, dx = -\int \frac{1}{u} \, du = -\log |u| + C = -\log |1 + \cos x| + C. \]

### Step 4: Solve the Third Integral
The third integral is:

\[ \int \frac{1}{x} \, dx = \log |x| + C. \]

### Step 5: Combine the Results
Now we combine all the results:

\[ \int \frac{1 + x \sin x + \cos x}{x(1 + \cos x)} \, dx = \log |x| - \log |1 + \cos x| + \log |x| + C. \]

### Final Step: Simplify the Expression
Combining the logarithmic terms gives:

\[ 2\log |x| - \log |1 + \cos x| + C = \log \left( \frac{x^2}{1 + \cos x} \right) + C. \]

Thus, the final answer is:

\[ \int \frac{1 + x \sin x + \cos x}{x(1 + \cos x)} \, dx = \log \left( \frac{x^2}{1 + \cos x} \right) + C. \]