Let `I_(n) = int_(0)^(1)x^(n)(tan^(1)x)dx, n in N`, then

Options

A. `(n+1)I_(n)+(n-1)I_(n-2)= (pi)/(4)+(1)/(n) AA n ge 3`
B. `(n+1)I_(n)+(n-1)I_(n-2)= (pi)/(2)-(1)/(n) AA n ge 3`
C. `(n+1)I_(n)-(n-1)I_(n-2)= (pi)/(2)+(1)/(n) AA n ge 3`
D. `(n+1)I_(n)-(n-1)I_(n-2)= (pi)/(2)-(1)/(n) AA n ge 3`

Answer

To solve the integral \( I_n = \int_0^1 x^n \tan^{-1}(x) \, dx \), where \( n \) is a natural number, we can use integration by parts. Here’s a step-by-step solution:

### Step 1: Set up the integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \tan^{-1}(x) \) (which is the inverse function)
- \( dv = x^n \, dx \) (which is the algebraic function)

### Step 2: Differentiate and integrate
Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{1+x^2} \, dx \]
- Integrate \( dv \):
\[ v = \int x^n \, dx = \frac{x^{n+1}}{n+1} \]

### Step 3: Apply the integration by parts formula
Substituting \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula gives:
\[ I_n = \left[ \tan^{-1}(x) \cdot \frac{x^{n+1}}{n+1} \right]_0^1 - \int_0^1 \frac{x^{n+1}}{n+1} \cdot \frac{1}{1+x^2} \, dx \]

### Step 4: Evaluate the boundary term
Evaluate the boundary term:
\[ \left[ \tan^{-1}(x) \cdot \frac{x^{n+1}}{n+1} \right]_0^1 = \tan^{-1}(1) \cdot \frac{1^{n+1}}{n+1} - \tan^{-1}(0) \cdot \frac{0^{n+1}}{n+1} = \frac{\pi}{4(n+1)} - 0 = \frac{\pi}{4(n+1)} \]

### Step 5: Simplify the integral
Now, we need to simplify the remaining integral:
\[ I_n = \frac{\pi}{4(n+1)} - \frac{1}{n+1} \int_0^1 \frac{x^{n+1}}{1+x^2} \, dx \]

### Step 6: Define the new integral
Let:
\[ J_n = \int_0^1 \frac{x^{n+1}}{1+x^2} \, dx \]
Thus, we have:
\[ I_n = \frac{\pi}{4(n+1)} - \frac{1}{n+1} J_n \]

### Step 7: Relate \( J_n \) back to \( I_n \)
To express \( J_n \) in terms of \( I_n \), we can use integration by parts again or find a relation. For simplicity, we can note that:
\[ J_n = \frac{1}{2} I_{n+1} \]
This gives us:
\[ I_n = \frac{\pi}{4(n+1)} - \frac{1}{2(n+1)} I_{n+1} \]

### Step 8: Solve the recurrence relation
We can now express \( I_n \) in terms of \( I_{n+1} \):
\[ (n+1) I_n + \frac{1}{2} I_{n+1} = \frac{\pi}{4} \]
This leads to a recurrence relation that can be solved iteratively.

### Final Result
The final expression for \( I_n \) can be derived from the recurrence relation and evaluated for specific values of \( n \).