If `I(m,n)=int_0^1t^m(1+t)^ndt,m,n,inR` , then I (m,n) is

Options

A. `n/(1+m)I["(m+1)",(n-1)]`
B. `m/(n+1)I["(m+1)",(n-1)]`
C. `2^n/(1+m)-n/(1+m)I["(m+1)",(n-1)]`
D. `2^n/(1+m)-m/(1+n)I["(m+1)",(n-1)]`

Answer

To solve the integral \( I(m,n) = \int_0^1 t^m (1+t)^n \, dt \), we will 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:
- \( u = (1+t)^n \) (which we will differentiate)
- \( dv = t^m \, dt \) (which we will integrate)

### Step 2: Differentiate and integrate
Now, we need to compute \( du \) and \( v \):
- Differentiate \( u \):
\[ du = n(1+t)^{n-1} \, dt \]
- Integrate \( dv \):
\[ v = \frac{t^{m+1}}{m+1} \]

### Step 3: Apply the integration by parts formula
Substituting into the integration by parts formula:
\[ I(m,n) = \left[ (1+t)^n \cdot \frac{t^{m+1}}{m+1} \right]_0^1 - \int_0^1 \frac{t^{m+1}}{m+1} \cdot n(1+t)^{n-1} \, dt \]

### Step 4: Evaluate the boundary terms
Now we evaluate the boundary terms:
- At \( t = 1 \):
\[ (1+1)^n \cdot \frac{1^{m+1}}{m+1} = \frac{2^n}{m+1} \]
- At \( t = 0 \):
\[ (1+0)^n \cdot \frac{0^{m+1}}{m+1} = 0 \]
Thus, the boundary terms contribute:
\[ \frac{2^n}{m+1} - 0 = \frac{2^n}{m+1} \]

### Step 5: Simplify the integral
Now we simplify the remaining integral:
\[ I(m,n) = \frac{2^n}{m+1} - \frac{n}{m+1} \int_0^1 t^{m+1} (1+t)^{n-1} \, dt \]
This integral can be rewritten as:
\[ I(m+1, n-1) = \int_0^1 t^{m+1} (1+t)^{n-1} \, dt \]
Thus, we have:
\[ I(m,n) = \frac{2^n}{m+1} - \frac{n}{m+1} I(m+1, n-1) \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ I(m,n) = \frac{2^n}{m+1} - \frac{n}{m+1} I(m+1, n-1) \]

### Final Result
The relationship can be expressed as:
\[ I(m,n) = \frac{2^n}{m+1} - \frac{n}{m+1} I(m+1, n-1) \]