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

Options

A. `(f(x))/(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 will use integration by parts. Let's break down the solution step by step.

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int [f(x) g''(x) - f''(x) g(x)] \, dx \]
This can be separated into two integrals:
\[ I = \int f(x) g''(x) \, dx - \int f''(x) g(x) \, dx \]

### Step 2: Apply Integration by Parts to the First Integral
For the first integral \( \int f(x) g''(x) \, dx \), we will use integration by parts. Let:
- \( u = f(x) \) (First function)
- \( dv = g''(x) \, dx \) (Second function)

Then, we differentiate and integrate:
- \( du = f'(x) \, dx \)
- \( v = g'(x) \)

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

### Step 3: Apply Integration by Parts to the Second Integral
Now, we consider the second integral \( \int f''(x) g(x) \, dx \). Again, we will use integration by parts. Let:
- \( u = g(x) \)
- \( dv = f''(x) \, dx \)

Then:
- \( du = g'(x) \, dx \)
- \( v = f'(x) \)

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

### Step 4: Combine the Results
Substituting the results from Steps 2 and 3 back into the expression for \( I \):
\[ I = \left( f(x) g'(x) - \int g'(x) f'(x) \, dx \right) - \left( g(x) f'(x) - \int f'(x) g'(x) \, dx \right) \]

This simplifies to:
\[ I = f(x) g'(x) - g(x) f'(x) - \int g'(x) f'(x) \, dx + \int f'(x) g'(x) \, dx \]

The last two integrals cancel each other out:
\[ I = f(x) g'(x) - g(x) f'(x) \]

### Step 5: Final Result
Thus, the final result of the integral is:
\[ \int [f(x) g''(x) - f''(x) g(x)] \, dx = f(x) g'(x) - g(x) f'(x) + C \]
where \( C \) is the constant of integration.