`int({x+sqrt(x^(2)+1)})^n/(sqrt(x^(2)+1))dx` is equal to

Options

A. `{x+sqrt(x^(2)+1)}^(n)+C`
B. `(1)/(n){x+sqrt(x^(2)+1)}^(n)+C`
C. `(1)/(n+1){x+sqrt(x^(2)+1)}^(n+1)+C`
D. none of these

Explanation

To solve the integral

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

we will follow these steps:

### Step 1: Rewrite the Integral
We can express the integral as:

\[ I = \int (x + \sqrt{x^2 + 1})^{n-1} (x + \sqrt{x^2 + 1}) \cdot \frac{1}{\sqrt{x^2 + 1}} \, dx. \]

### Step 2: Substitution
Let

\[ t = x + \sqrt{x^2 + 1}. \]

Now, we need to find \( dt \). Differentiating both sides with respect to \( x \):

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

This simplifies to:

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

Thus, we can express \( dx \) in terms of \( dt \):

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

### Step 3: Substitute Back into the Integral
Now substituting \( t \) and \( dx \) into the integral:

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

### Step 4: Integrate
Now we can integrate:

\[ I = \int t^{n-1} \, dt = \frac{t^n}{n} + C, \]

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

### Step 5: Substitute Back for \( t \)
Finally, substitute back the value of \( t \):

\[ I = \frac{(x + \sqrt{x^2 + 1})^n}{n} + C. \]

### Final Answer
Thus, the integral evaluates to:

\[ \int \frac{(x + \sqrt{x^2 + 1})^n}{\sqrt{x^2 + 1}} \, dx = \frac{(x + \sqrt{x^2 + 1})^n}{n} + C. \]

---