`int(cosx+xsinx)/(x^(2)+xcosx)dx= ` . . . .

Options

A. `log(x(x+cosx))+C`
B. `log((x)/(x+cosx))+C`
C. `log((x+cosx)/(x))`
D. none of these

Answer

To solve the integral

\[ I = \int \frac{\cos x + x \sin x}{x^2 + x \cos x} \, dx, \]

we will follow a systematic approach.

### Step 1: Rewrite the Integral
We start by rewriting the integral \( I \):

\[ I = \int \frac{\cos x + x \sin x}{x^2 + x \cos x} \, dx. \]

### Step 2: Factor the Numerator
We can factor out \( x \) from the numerator:

\[ I = \int \frac{x(x + \cos x) + (\cos x)}{x^2 + x \cos x} \, dx. \]

### Step 3: Split the Numerator
Next, we will split the numerator into two parts:

\[ I = \int \left( \frac{x + \cos x}{x(x + \cos x)} + \frac{\sin x - 1}{x + \cos x} \right) \, dx. \]

### Step 4: Separate the Integrals
Now we can separate the integrals:

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

### Step 5: Solve the First Integral
The first integral is straightforward:

\[ \int \frac{1}{x} \, dx = \ln |x| + C_1. \]

### Step 6: Solve the Second Integral
For the second integral, we will use substitution. Let

\[ t = x + \cos x. \]

Then, differentiating gives us:

\[ dt = (1 - \sin x) \, dx. \]

We can express \( dx \) in terms of \( dt \):

\[ dx = \frac{dt}{1 - \sin x}. \]

Now, we rewrite the integral:

\[ \int \frac{\sin x - 1}{t} \, \frac{dt}{1 - \sin x}. \]

This simplifies to:

\[ -\int \frac{1}{t} \, dt = -\ln |t| + C_2 = -\ln |x + \cos x| + C_2. \]

### Step 7: Combine the Results
Now we combine both integrals:

\[ I = \ln |x| - \ln |x + \cos x| + C, \]

where \( C = C_1 + C_2 \).

### Step 8: Final Result
Using the properties of logarithms, we can combine the logarithms:

\[ I = \ln \left| \frac{x}{x + \cos x} \right| + C. \]

Thus, the final result for the integral is:

\[ \int \frac{\cos x + x \sin x}{x^2 + x \cos x} \, dx = \ln \left| \frac{x}{x + \cos x} \right| + C. \]

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