Given that `int (dx)/((1+ x ^(2))^n)=(x)/(2 (n-1)(1+x^(2) )^(n-1))+((2n -3))/(2(n-1))int (dx)/((1+ x ^(2))^(n-1)).` Find the vlaue of `int _(0)^(1) (dx)/((1+x ^(2) )^(4)),` (you may or may not use reduction formula given)

Options

A. 1: `11/48+(5pi)/(64)`
B. 2: `(11)/(48)+ (5pi)/(32)`
C. 3: `(1)/(24) + (5pi)/(64)`
D. 4: `(1)/(96) + (5pi)/(32)`

Answer

To solve the integral \( \int_0^1 \frac{dx}{(1+x^2)^4} \) using the provided formula, we will follow these steps:

### Step 1: Identify the parameters
We are given the formula:
\[ \int \frac{dx}{(1+x^2)^n} = \frac{x}{2(n-1)(1+x^2)^{n-1}} + \frac{2n-3}{2(n-1)} \int \frac{dx}{(1+x^2)^{n-1}} \]
For our case, we set \( n = 4 \).

### Step 2: Apply the formula
Substituting \( n = 4 \) into the formula:
\[ \int \frac{dx}{(1+x^2)^4} = \frac{x}{2(4-1)(1+x^2)^{3}} + \frac{2(4)-3}{2(4-1)} \int \frac{dx}{(1+x^2)^{3}} \]
This simplifies to:
\[ \int \frac{dx}{(1+x^2)^4} = \frac{x}{6(1+x^2)^{3}} + \frac{5}{6} \int \frac{dx}{(1+x^2)^{3}} \]

### Step 3: Evaluate the integral \( \int \frac{dx}{(1+x^2)^3} \)
We apply the formula again for \( n = 3 \):
\[ \int \frac{dx}{(1+x^2)^3} = \frac{x}{2(3-1)(1+x^2)^{2}} + \frac{2(3)-3}{2(3-1)} \int \frac{dx}{(1+x^2)^{2}} \]
This simplifies to:
\[ \int \frac{dx}{(1+x^2)^3} = \frac{x}{4(1+x^2)^{2}} + \frac{3}{4} \int \frac{dx}{(1+x^2)^{2}} \]

### Step 4: Evaluate the integral \( \int \frac{dx}{(1+x^2)^2} \)
We apply the formula again for \( n = 2 \):
\[ \int \frac{dx}{(1+x^2)^2} = \frac{x}{2(2-1)(1+x^2)^{1}} + \frac{2(2)-3}{2(2-1)} \int \frac{dx}{(1+x^2)^{1}} \]
This simplifies to:
\[ \int \frac{dx}{(1+x^2)^2} = \frac{x}{2(1+x^2)} + \frac{1}{2} \int \frac{dx}{(1+x^2)} \]

### Step 5: Evaluate the integral \( \int \frac{dx}{(1+x^2)} \)
This integral is known:
\[ \int \frac{dx}{(1+x^2)} = \tan^{-1}(x) \]

### Step 6: Substitute back
Now we can substitute back into our previous integrals:
1. For \( \int \frac{dx}{(1+x^2)^2} \):
\[ \int \frac{dx}{(1+x^2)^2} = \frac{x}{2(1+x^2)} + \frac{1}{2} \tan^{-1}(x) \]

2. For \( \int \frac{dx}{(1+x^2)^3} \):
\[ \int \frac{dx}{(1+x^2)^3} = \frac{x}{4(1+x^2)^{2}} + \frac{3}{4} \left( \frac{x}{2(1+x^2)} + \frac{1}{2} \tan^{-1}(x) \right) \]

3. For \( \int \frac{dx}{(1+x^2)^4} \):
\[ \int \frac{dx}{(1+x^2)^4} = \frac{x}{6(1+x^2)^{3}} + \frac{5}{6} \left( \frac{x}{4(1+x^2)^{2}} + \frac{3}{4} \left( \frac{x}{2(1+x^2)} + \frac{1}{2} \tan^{-1}(x) \right) \right) \]

### Step 7: Evaluate from 0 to 1
Now we need to evaluate:
\[ \int_0^1 \frac{dx}{(1+x^2)^4} \]
Substituting the limits into the expression we derived for \( \int \frac{dx}{(1+x^2)^4} \).

### Step 8: Final Calculation
After substituting \( x = 1 \) and \( x = 0 \) into the final expression and simplifying, we will arrive at the final answer.