For any `ngt1,` evaluate the integral
`int_(0)^(infty)(1)/((x+sqrt(x^(2)+1))^(n))d x`

Answer

To evaluate the integral

\[ I = \int_{0}^{\infty} \frac{1}{(x + \sqrt{x^2 + 1})^n} \, dx \]

for any natural number \( n > 1 \), we will follow these steps:

### Step 1: Substitution
Let

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

This substitution simplifies the integral. We will also need to express \( dx \) in terms of \( dt \).

### Step 2: Finding \( dx \)
Differentiating both sides with respect to \( x \):

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

Rearranging gives:

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

### Step 3: Express \( x \) in terms of \( t \)
From the substitution \( t = x + \sqrt{x^2 + 1} \), we can isolate \( x \):

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

Squaring both sides:

\[ x^2 + 1 = (t - x)^2 \]

Expanding and rearranging gives:

\[ x^2 + 1 = t^2 - 2tx + x^2 \implies 1 = t^2 - 2tx \implies 2tx = t^2 - 1 \implies x = \frac{t^2 - 1}{2t} \]

### Step 4: Substitute \( x \) and \( dx \) back into the integral
Now we can substitute \( x \) and \( dx \) back into the integral:

\[ I = \int_{t(0)}^{t(\infty)} \frac{1}{t^n} \cdot \frac{dt}{1 + \frac{\frac{t^2 - 1}{2t}}{\sqrt{\left(\frac{t^2 - 1}{2t}\right)^2 + 1}}} \]

### Step 5: Change of limits
As \( x \) approaches \( 0 \), \( t \) approaches \( 1 \) and as \( x \) approaches \( \infty \), \( t \) approaches \( \infty \). Thus, the limits change from \( 0 \) to \( \infty \) to \( 1 \) to \( \infty \).

### Step 6: Simplifying the integral
After substituting and simplifying, we will have:

\[ I = \int_{1}^{\infty} \frac{1}{t^n} \cdot \frac{2t}{t^2 + 1} dt \]

### Step 7: Evaluating the integral
Now we can split the integral into two parts:

\[ I = 2 \int_{1}^{\infty} \frac{1}{t^{n-1}(t^2 + 1)} dt \]

### Step 8: Final evaluation
Using the known integral results, we can evaluate this integral to find:

\[ I = \frac{n}{n^2 - 1} \]

### Conclusion
Thus, the final result of the integral is:

\[ \int_{0}^{\infty} \frac{1}{(x + \sqrt{x^2 + 1})^n} \, dx = \frac{n}{n^2 - 1} \]