If `int g(x) dx = g (x) ` , then `int(f(x) + f' (x)) g (x) dx` is equal to

Options

A. `f(x) g (x)`
B. `f'(x) g (x)`
C. `(f(x) -f'(x))g(x)`
D. `f(x)g'(x)`

Answer

To solve the problem, we need to evaluate the integral \( \int (f(x) + f'(x)) g(x) \, dx \) given that \( \int g(x) \, dx = g(x) \).

### Step-by-Step Solution:

1. **Rewrite the Integral**:
We start with the integral:
\[ \int (f(x) + f'(x)) g(x) \, dx \]
This can be split into two separate integrals:
\[ \int f(x) g(x) \, dx + \int f'(x) g(x) \, dx \]

2. **Apply Integration by Parts**:
For the second integral \( \int f'(x) g(x) \, dx \), we will use integration by parts. Recall the formula:
\[ \int u \, dv = uv - \int v \, du \]
Let \( u = f(x) \) and \( dv = g(x) \, dx \). Then, we have:
- \( du = f'(x) \, dx \)
- \( v = \int g(x) \, dx = g(x) \) (given)

Applying integration by parts:
\[ \int f'(x) g(x) \, dx = f(x) g(x) - \int f(x) g'(x) \, dx \]

3. **Combine the Integrals**:
Now we can substitute this back into our expression:
\[ \int (f(x) + f'(x)) g(x) \, dx = \int f(x) g(x) \, dx + \left( f(x) g(x) - \int f(x) g'(x) \, dx \right) \]
Simplifying this gives:
\[ \int (f(x) + f'(x)) g(x) \, dx = f(x) g(x) + \int f(x) g(x) \, dx - \int f(x) g'(x) \, dx \]

4. **Simplify Further**:
Notice that \( \int f(x) g(x) \, dx \) appears on both sides. We can rearrange the equation:
\[ \int (f(x) + f'(x)) g(x) \, dx - \int f(x) g(x) \, dx = f(x) g(x) - \int f(x) g'(x) \, dx \]
This implies:
\[ \int f(x) g(x) \, dx = f(x) g(x) + C \]
where \( C \) is a constant of integration.

5. **Final Result**:
Thus, we conclude:
\[ \int (f(x) + f'(x)) g(x) \, dx = f(x) g(x) \]

### Final Answer:
The integral \( \int (f(x) + f'(x)) g(x) \, dx \) is equal to \( f(x) g(x) \).