If `n gt 1`. Evaluate `int_(0)^(oo)(dx)/((x+sqrt(1+x^(2)))^(n))`

Answer

To evaluate the integral

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

where \( n > 1 \), we can follow these steps:

### Step 1: Substitution
Let \( p = x + \sqrt{1 + x^2} \).

From this, we can express \( \sqrt{1 + x^2} \) in terms of \( p \):
\[ \sqrt{1 + x^2} = p - x \]

### Step 2: Squaring Both Sides
Squaring both sides gives:
\[ 1 + x^2 = (p - x)^2 = p^2 - 2px + x^2 \]
This simplifies to:
\[ 1 = p^2 - 2px \implies 2px = p^2 - 1 \implies x = \frac{p^2 - 1}{2p} \]

### Step 3: Finding \( dx \)
Now we differentiate \( x \) with respect to \( p \):
\[ dx = \frac{d}{dp}\left(\frac{p^2 - 1}{2p}\right) dp \]
Using the quotient rule:
\[ dx = \left(\frac{(2p)(2p) - (p^2 - 1)(2)}{(2p)^2}\right) dp = \frac{4p^2 - 2(p^2 - 1)}{4p^2} dp = \frac{2p^2 + 2}{4p^2} dp = \frac{p^2 + 1}{2p^2} dp \]

### Step 4: Changing the Limits of Integration
When \( x = 0 \):
\[ p = 0 + \sqrt{1 + 0^2} = 1 \]
When \( x \to \infty \):
\[ p = \infty + \sqrt{1 + \infty^2} = \infty \]

### Step 5: Substitute in the Integral
Now substitute \( x \) and \( dx \) into the integral:
\[ I = \int_{1}^{\infty} \frac{p^2 + 1}{2p^2} \cdot \frac{1}{p^n} dp \]
This simplifies to:
\[ I = \frac{1}{2} \int_{1}^{\infty} \left( \frac{p^2 + 1}{p^n} \right) dp = \frac{1}{2} \left( \int_{1}^{\infty} \frac{p^2}{p^n} dp + \int_{1}^{\infty} \frac{1}{p^n} dp \right) \]

### Step 6: Evaluate Each Integral
1. For the first integral:
\[ \int_{1}^{\infty} \frac{p^2}{p^n} dp = \int_{1}^{\infty} p^{2-n} dp \]
This converges for \( n > 3 \) and evaluates to:
\[ \frac{1}{3-n} \quad \text{(for \( n < 3 \))} \]

2. For the second integral:
\[ \int_{1}^{\infty} \frac{1}{p^n} dp = \frac{1}{n-1} \quad \text{(for \( n > 1 \))} \]

### Step 7: Combine Results
Thus, we have:
\[ I = \frac{1}{2} \left( \frac{1}{3-n} + \frac{1}{n-1} \right) \]

### Final Result
The final result for the integral is:
\[ I = \frac{1}{2} \left( \frac{1}{3-n} + \frac{1}{n-1} \right) \]