`int(cosx-sinx+1-x)/(e^(x)+sinx+x)dx=log_(e)(f(x))+g(x)+C` where C is the constant of integration and f(x) is positive. Then `f(x)+g(x)` has the value equal to

Options

A. `e^(x)+sinx+2x`
B. `e^(x)+sinx`
C. `e^(x)-sinx`
D. `e^(x)+sinx+x`

Explanation

To solve the integral

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

we can manipulate the integrand to make it easier to integrate.

### Step 1: Rewrite the integrand
We can rewrite the numerator by adding and subtracting \( e^x \):

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

This gives us:

\[ I = \int \frac{(\cos x + 1 + e^x) - (\sin x + e^x + x)}{e^x + \sin x + x} \, dx \]

### Step 2: Separate the integral
Now we can separate the integral into two parts:

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

### Step 3: Let \( t = e^x + \sin x + x \)
Next, we can use substitution. Let:

\[ t = e^x + \sin x + x \]

Then, the differential \( dt \) is:

\[ dt = (e^x + \cos x + 1) \, dx \]

Thus, we can express \( dx \) in terms of \( dt \):

\[ dx = \frac{dt}{e^x + \cos x + 1} \]

### Step 4: Substitute back into the integral
Now, substituting back into the integral:

\[ I = \int \frac{1}{t} \, dt - \int 1 \, dx \]

The first integral becomes:

\[ \int \frac{1}{t} \, dt = \ln |t| + C \]

And the second integral is simply:

\[ -x \]

### Step 5: Combine the results
Combining these results, we have:

\[ I = \ln |e^x + \sin x + x| - x + C \]

### Step 6: Identify \( f(x) \) and \( g(x) \)
From the given form \( I = \ln(f(x)) + g(x) + C \), we can identify:

\[ f(x) = e^x + \sin x + x \]
\[ g(x) = -x \]

### Step 7: Find \( f(x) + g(x) \)
Now, we need to find \( f(x) + g(x) \):

\[ f(x) + g(x) = (e^x + \sin x + x) + (-x) = e^x + \sin x \]

### Final Answer
Thus, the value of \( f(x) + g(x) \) is:

\[ \boxed{e^x + \sin x} \]