If `f(x) = ln (x - sqrt(1 + x^(2)))`, then what is `int f''(x)` dx equal to ?

Options

A. `(1)/((x - sqrt(1 + x^(2)))) + c`
B. `-(1)/(sqrt(1 + x^(2))) + c`
C. `-sqrt(1 + x^(2)) + c`
D. `ln (x - sqrt(1+x^(2))) + c`

Explanation

To solve the problem, we need to find the integral of the second derivative of the function \( f(x) = \ln(x - \sqrt{1 + x^2}) \). We will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)

Using the chain rule and the properties of logarithmic differentiation, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \ln(x - \sqrt{1 + x^2}) \right) \]

Using the derivative of \( \ln(u) \) which is \( \frac{1}{u} \cdot \frac{du}{dx} \):

Let \( u = x - \sqrt{1 + x^2} \).

Then,

\[ \frac{du}{dx} = 1 - \frac{x}{\sqrt{1 + x^2}} \]

Thus,

\[ f'(x) = \frac{1}{x - \sqrt{1 + x^2}} \cdot \left( 1 - \frac{x}{\sqrt{1 + x^2}} \right) \]

### Step 2: Simplify \( f'(x) \)

Now we simplify \( f'(x) \):

\[ f'(x) = \frac{1 - \frac{x}{\sqrt{1 + x^2}}}{x - \sqrt{1 + x^2}} \]

This can be further simplified, but for our purpose, we will proceed to find the second derivative.

### Step 3: Find the second derivative \( f''(x) \)

Now we differentiate \( f'(x) \) to find \( f''(x) \):

Using the quotient rule:

\[ f''(x) = \frac{(v \cdot u' - u \cdot v')}{v^2} \]

Where \( u = 1 - \frac{x}{\sqrt{1 + x^2}} \) and \( v = x - \sqrt{1 + x^2} \).

Calculating \( u' \) and \( v' \):

1. \( u' = -\frac{\sqrt{1 + x^2} - \frac{x^2}{\sqrt{1 + x^2}}}{1 + x^2} = -\frac{1}{(1 + x^2)^{3/2}} \)
2. \( v' = 1 - \frac{x}{\sqrt{1 + x^2}} \)

Substituting these into the quotient rule gives us \( f''(x) \).

### Step 4: Integrate \( f''(x) \)

Now we need to integrate \( f''(x) \):

\[ \int f''(x) \, dx = f'(x) + C \]

Where \( C \) is the constant of integration.

### Step 5: Substitute back to find \( f'(x) \)

Since we have already derived \( f'(x) \), we can express our final answer:

\[ \int f''(x) \, dx = f'(x) + C = \frac{1 - \frac{x}{\sqrt{1 + x^2}}}{x - \sqrt{1 + x^2}} + C \]

### Final Answer

Thus,

\[ \int f''(x) \, dx = f'(x) + C \]