Integrating by parts, derive reduction formulas for calculating the following integrals :
(a) `I_(n) = int (dx)/( (x^(2) + a^(2) )^(n) )`,

Answer

To derive the reduction formula for the integral

\[ I_n = \int \frac{dx}{(x^2 + a^2)^n}, \]

we will use the method of integration by parts. Let's go through the steps systematically.

### Step 1: Choose functions for integration by parts

We will use the integration by parts formula:

\[ \int u \, dv = uv - \int v \, du. \]

Let:
- \( u = \frac{1}{(x^2 + a^2)^{n-1}} \) (which will be differentiated),
- \( dv = \frac{dx}{(x^2 + a^2)} \) (which will be integrated).

### Step 2: Differentiate \( u \) and integrate \( dv \)

Now we need to find \( du \) and \( v \):

1. Differentiate \( u \):
\[ du = -\frac{(n-1) \cdot 2x}{(x^2 + a^2)^{n}} \, dx = -\frac{2(n-1)x}{(x^2 + a^2)^{n}} \, dx. \]

2. Integrate \( dv \):
\[ v = \int \frac{dx}{(x^2 + a^2)} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right). \]

### Step 3: Apply integration by parts

Now we apply the integration by parts formula:

\[ I_n = uv - \int v \, du. \]

Substituting \( u \), \( v \), and \( du \):

\[ I_n = \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) - \int \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \left(-\frac{2(n-1)x}{(x^2 + a^2)^{n}}\right) \, dx. \]

### Step 4: Simplifying the integral

This gives us:

\[ I_n = \frac{1}{a(x^2 + a^2)^{n-1}} \tan^{-1}\left(\frac{x}{a}\right) + \frac{2(n-1)}{a} \int \frac{x \tan^{-1}\left(\frac{x}{a}\right)}{(x^2 + a^2)^{n}} \, dx. \]

### Step 5: Rearranging to find the reduction formula

Now, we can rearrange the terms to isolate \( I_n \):

\[ I_n = \frac{1}{a(x^2 + a^2)^{n-1}} \tan^{-1}\left(\frac{x}{a}\right) + \frac{2(n-1)}{a} I_{n-1}. \]

### Step 6: Final reduction formula

Thus, we have derived the reduction formula:

\[ I_n = \frac{1}{a(x^2 + a^2)^{n-1}} \tan^{-1}\left(\frac{x}{a}\right) + \frac{2(n-1)}{a} I_{n-1}. \]