Let `f:DtoR`, where D is the domain of `f`. Find the inverse of `f` if it exists: `f(x)=ln(x+sqrt(1+x^(2)))`

Explanation

To find the inverse of the function \( f(x) = \ln(x + \sqrt{1+x^2}) \), we will follow these steps:

### Step 1: Replace \( f(x) \) with \( y \)
Let \( y = f(x) \). Therefore, we have:
\[ y = \ln(x + \sqrt{1+x^2}) \]

### Step 2: Swap \( x \) and \( y \)
To find the inverse, we swap \( x \) and \( y \):
\[ x = \ln(y + \sqrt{1+y^2}) \]

### Step 3: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides:
\[ e^x = y + \sqrt{1+y^2} \]

### Step 4: Isolate the square root
Now, we isolate the square root:
\[ \sqrt{1+y^2} = e^x - y \]

### Step 5: Square both sides
Next, we square both sides to eliminate the square root:
\[ 1 + y^2 = (e^x - y)^2 \]

### Step 6: Expand the right side
Expanding the right side gives:
\[ 1 + y^2 = e^{2x} - 2ye^x + y^2 \]

### Step 7: Simplify the equation
Now, we can simplify the equation by canceling \( y^2 \) from both sides:
\[ 1 = e^{2x} - 2ye^x \]

### Step 8: Solve for \( y \)
Rearranging the equation to solve for \( y \):
\[ 2ye^x = e^{2x} - 1 \]
\[ y = \frac{e^{2x} - 1}{2e^x} \]

### Step 9: Replace \( y \) with \( f^{-1}(x) \)
Thus, the inverse function is:
\[ f^{-1}(x) = \frac{e^{2x} - 1}{2e^x} \]

### Final Answer
The inverse of the function \( f(x) = \ln(x + \sqrt{1+x^2}) \) is:
\[ f^{-1}(x) = \frac{e^{2x} - 1}{2e^x} \]