To evaluate the integral
\[
I = \int_0^{\infty} \frac{dx}{(x + \sqrt{1 + x^2})^n}
\]
where \( n > 1 \) is an integer, we will use the substitution \( x = \tan \theta \).
### Step 1: Change of Variables
Using the substitution \( x = \tan \theta \), we have:
\[
dx = \sec^2 \theta \, d\theta
\]
The limits change as follows:
- When \( x = 0 \), \( \theta = 0 \).
- When \( x \to \infty \), \( \theta \to \frac{\pi}{2} \).
Thus, the integral becomes:
\[
I = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta \, d\theta}{(\tan \theta + \sqrt{1 + \tan^2 \theta})^n}
\]
### Step 2: Simplifying the Integrand
Using the identity \( \sqrt{1 + \tan^2 \theta} = \sec \theta \), we can rewrite the integrand:
\[
I = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta \, d\theta}{(\tan \theta + \sec \theta)^n}
\]
### Step 3: Further Simplification
We can express \( \tan \theta + \sec \theta \) in a more manageable form. Let:
\[
z = \tan \theta + \sec \theta
\]
Then, we differentiate:
\[
dz = (\sec^2 \theta + \sec \theta \tan \theta) d\theta
\]
This means:
\[
d\theta = \frac{dz}{\sec^2 \theta + \sec \theta \tan \theta}
\]
### Step 4: Expressing in Terms of \( z \)
Now, we need to express \( \sec^2 \theta \) in terms of \( z \). From the identity \( \sec^2 \theta - \tan^2 \theta = 1 \), we can derive:
\[
\sec^2 \theta = z^2 - 1
\]
Thus, substituting back into the integral, we have:
\[
I = \int_1^{\infty} \frac{(z^2 - 1) \, dz}{z^n}
\]
### Step 5: Splitting the Integral
Now we split the integral:
\[
I = \int_1^{\infty} (z^{2-n} - z^{-n}) \, dz
\]
### Step 6: Evaluating the Integrals
Now we evaluate each integral separately:
1. For \( \int_1^{\infty} z^{2-n} \, dz \):
\[
\int_1^{\infty} z^{2-n} \, dz = \left[ \frac{z^{3-n}}{3-n} \right]_1^{\infty}
\]
This converges if \( n > 1 \) (which it does), giving:
\[
= \frac{1}{n-3} \quad \text{(for } n < 3\text{)}
\]
2. For \( \int_1^{\infty} z^{-n} \, dz \):
\[
\int_1^{\infty} z^{-n} \, dz = \left[ \frac{z^{1-n}}{1-n} \right]_1^{\infty}
\]
This converges if \( n > 1 \), giving:
\[
= \frac{1}{n-1}
\]
### Step 7: Combining Results
Thus, we have:
\[
I = \frac{1}{3-n} - \frac{1}{n-1}
\]
### Final Result
Combining these results gives us:
\[
I = \frac{1}{1 - n} - \frac{1}{3 - n}
\]
This simplifies to:
\[
I = \frac{2}{(n-1)(3-n)}
\]
