If `int(log (x + sqrt(1 + x^2)))/(sqrt(1 + x^2)) dx = (fog) x +c` then , the function f and g are respectively

Options

A. `log sqrt(1 + x^2) + x`
B. `log (x + sqrt(1 + x^2)) x^2`
C. `(x^2)/2 ,"log" (x + sqrt(1 + x^2))`
D. none

Answer

To solve the given integral equation:

\[ \int \frac{\log(x + \sqrt{1 + x^2})}{\sqrt{1 + x^2}} \, dx = f(g(x)) + c \]

we will follow these steps:

### Step 1: Substitution
Let:
\[ t = x + \sqrt{1 + x^2} \]

### Step 2: Differentiate to find \(dx\)
To find \(dx\), we differentiate \(t\):
\[ dt = \left(1 + \frac{x}{\sqrt{1 + x^2}}\right) dx \]
This simplifies to:
\[ dt = \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}} \, dx \]
Thus, we have:
\[ dx = \frac{\sqrt{1 + x^2}}{\sqrt{1 + x^2} + x} \, dt \]

### Step 3: Rewrite the integral
Now substitute \(t\) and \(dx\) into the integral:
\[ \int \frac{\log(t)}{\sqrt{1 + x^2}} \cdot \frac{\sqrt{1 + x^2}}{\sqrt{1 + x^2} + x} \, dt \]
This simplifies to:
\[ \int \frac{\log(t)}{t} \, dt \]

### Step 4: Integrate
The integral of \(\frac{\log(t)}{t}\) is:
\[ \frac{(\log(t))^2}{2} + C \]

### Step 5: Substitute back for \(t\)
Now we substitute back \(t = x + \sqrt{1 + x^2}\):
\[ \frac{(\log(x + \sqrt{1 + x^2}))^2}{2} + C \]

### Step 6: Identify \(f\) and \(g\)
From the expression:
\[ f(g(x)) = \frac{(\log(x + \sqrt{1 + x^2}))^2}{2} \]
We can identify:
- \(g(x) = x + \sqrt{1 + x^2}\)
- \(f(x) = \frac{x^2}{2}\)

### Final Answer
Thus, the functions \(f\) and \(g\) are:
- \(f(x) = \frac{x^2}{2}\)
- \(g(x) = x + \sqrt{1 + x^2}\)