If `int(log"("x+sqrt(1+x^(2))")")/(sqrt(1+x^(2)))dx="gof"(x)+"constant, then"`

Options

A. `f(x)=log(x+sqrt(x^(2)+1))`
B. `f(x)=log(x+sqrt(x^(2)+1))" and "g(x)=x^(2)`
C. `f(x)=log(x+sqrt(x^(2)+1))" and "g(x)=(x^(2))/(2)`
D. `f(x)=(x^(2))/(2)" and "g(x)=log(x+sqrt(x^(2)+1))`

Explanation

To solve the problem, we need to evaluate the integral

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

and express it in the form \( g(f(x)) + C \).

### Step 1: Substitution
Let

\[ t = \log(x + \sqrt{1 + x^2}) \]

We will differentiate \( t \) with respect to \( x \).

**Hint:** Use the chain rule for differentiation.

### Step 2: Differentiate \( t \)
The derivative of \( t \) is given by:

\[ \frac{dt}{dx} = \frac{1}{x + \sqrt{1 + x^2}} \cdot \left(1 + \frac{x}{\sqrt{1 + x^2}}\right) \]

This simplifies to:

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

**Hint:** Simplify the expression by finding a common denominator.

### Step 3: Simplify the Derivative
The expression simplifies to:

\[ \frac{dt}{dx} = \frac{\sqrt{1 + x^2} + x}{x + \sqrt{1 + x^2}} \cdot \frac{1}{\sqrt{1 + x^2}} = \frac{1}{\sqrt{1 + x^2}} \]

This means:

\[ dx = \sqrt{1 + x^2} \, dt \]

**Hint:** Substitute \( dx \) back into the integral.

### Step 4: Substitute in the Integral
Now, substituting \( dx \) into the integral:

\[ \int \frac{t}{\sqrt{1 + x^2}} \cdot \sqrt{1 + x^2} \, dt = \int t \, dt \]

**Hint:** Recognize that this integral is straightforward.

### Step 5: Integrate
Integrating \( t \):

\[ \int t \, dt = \frac{t^2}{2} + C \]

**Hint:** Remember to substitute back for \( t \).

### Step 6: Substitute Back
Substituting \( t \) back into the equation gives:

\[ \frac{1}{2} \left(\log(x + \sqrt{1 + x^2})\right)^2 + C \]

**Hint:** Identify \( f(x) \) and \( g(x) \) from the expression.

### Step 7: Identify \( f(x) \) and \( g(x) \)
From the integral result, we can identify:

- \( f(x) = \log(x + \sqrt{1 + x^2}) \)
- \( g(x) = \frac{x^2}{2} \)

Thus, we have:

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

### Final Answer
The values of \( f(x) \) and \( g(x) \) are:

- \( f(x) = \log(x + \sqrt{1 + x^2}) \)
- \( g(x) = \frac{x^2}{2} \)