The value of `int [f(x)g''(x) - f''(x)g(x)] dx` is equal to

Options

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

Answer

To solve the integral \( \int [f(x) g''(x) - f''(x) g(x)] \, dx \), we can use integration by parts. Here’s a step-by-step solution:

### Step 1: Identify the Functions
We rewrite the integral:
\[ I = \int [f(x) g''(x) - f''(x) g(x)] \, dx \]
We can split this into two separate integrals:
\[ I = \int f(x) g''(x) \, dx - \int f''(x) g(x) \, dx \]

### Step 2: Apply Integration by Parts
For the first integral \( \int f(x) g''(x) \, dx \), we can use integration by parts. Let:
- \( u = f(x) \) → \( du = f'(x) \, dx \)
- \( dv = g''(x) \, dx \) → \( v = g'(x) \)

Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int f(x) g''(x) \, dx = f(x) g'(x) - \int g'(x) f'(x) \, dx \]

### Step 3: Substitute Back
Now substituting this back into our expression for \( I \):
\[ I = \left[ f(x) g'(x) - \int g'(x) f'(x) \, dx \right] - \int f''(x) g(x) \, dx \]

### Step 4: Combine the Integrals
Now we need to handle the second integral \( -\int f''(x) g(x) \, dx \). We can again apply integration by parts:
Let:
- \( u = g(x) \) → \( du = g'(x) \, dx \)
- \( dv = f''(x) \, dx \) → \( v = f'(x) \)

Thus:
\[ -\int f''(x) g(x) \, dx = -\left[ g(x) f'(x) - \int f'(x) g'(x) \, dx \right] \]

### Step 5: Combine Everything
Now, substituting this back into our expression for \( I \):
\[ I = f(x) g'(x) - \int g'(x) f'(x) \, dx - g(x) f'(x) + \int f'(x) g'(x) \, dx \]
Notice that the terms \( -\int g'(x) f'(x) \, dx \) and \( \int f'(x) g'(x) \, dx \) cancel each other out:
\[ I = f(x) g'(x) - g(x) f'(x) \]

### Final Result
Thus, the value of the integral is:
\[ \int [f(x) g''(x) - f''(x) g(x)] \, dx = f(x) g'(x) - g(x) f'(x) \]