Evaluate ` l_(n)= int (dx)/((x^(2)+a^(2))^(n))`.

Answer

To evaluate the integral \( I_n = \int \frac{dx}{(x^2 + a^2)^n} \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
\[ I_n = \int \frac{dx}{(x^2 + a^2)^n} \]

2. **Multiply and Divide by \( a^2 \)**:
\[ I_n = \frac{1}{a^2} \int \frac{a^2}{(x^2 + a^2)^n} \, dx \]

3. **Add and Subtract \( x^2 \)**:
\[ I_n = \frac{1}{a^2} \left( \int \frac{x^2 + a^2}{(x^2 + a^2)^n} \, dx - \int \frac{x^2}{(x^2 + a^2)^n} \, dx \right) \]

4. **Separate the Integrals**:
\[ I_n = \frac{1}{a^2} \left( \int \frac{dx}{(x^2 + a^2)^{n-1}} - \int \frac{x^2 \, dx}{(x^2 + a^2)^n} \right) \]

5. **Substitute for the Second Integral**:
For the second integral, we can use the substitution \( t = x^2 + a^2 \), which gives \( dt = 2x \, dx \) or \( dx = \frac{dt}{2x} \). Thus, we have:
\[ \int \frac{x^2 \, dx}{(x^2 + a^2)^n} = \int \frac{x^2}{t^n} \cdot \frac{dt}{2x} = \frac{1}{2} \int \frac{x}{t^n} \, dt \]

6. **Evaluate the Integral**:
The integral can be evaluated using integration techniques, but we will focus on the recursive relationship. We can express \( I_n \) in terms of \( I_{n-1} \):
\[ I_n = \frac{1}{a^2} \left( I_{n-1} - \frac{1}{2(n-1)} \cdot \frac{1}{(x^2 + a^2)^{n-1}} \right) \]

7. **Establish the Recursive Formula**:
After simplifying, we can derive a recursive formula:
\[ I_{n+1} = \frac{2n - 1}{2n} \cdot \frac{1}{a^2} I_n + \frac{x}{2n} \cdot \frac{1}{(x^2 + a^2)^{n}} \]

### Final Result:
Thus, we have established a recursive relationship for \( I_n \):
\[ I_{n+1} = \frac{2n - 1}{2n a^2} I_n + \frac{x}{2n a^2 (x^2 + a^2)^{n}} \]