If `I_(m,n)=overset1underset0intx^(m-1)(1-x)^(n-1)dx` for `m,nge1` and `overset1underset0int(x^(m-1)+x^(n-1))/((1+x)^(m+n))dx=alphaI_(m,n),alpha in R` then `alpha` equals `"_____"`

Answer

To solve the problem, we need to evaluate the integral

\[ \int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}} \, dx \]

and express it in terms of \( I_{m,n} \), which is defined as:

\[ I_{m,n} = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx \]

### Step 1: Rewrite the Integral

We start with the given integral:

\[ \int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}} \, dx \]

We can separate this into two integrals:

\[ \int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}} \, dx + \int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}} \, dx \]

### Step 2: Use Substitution

For each integral, we can use the substitution \( x = \frac{y}{1-y} \), which gives \( dx = \frac{1}{(1-y)^2} \, dy \). The limits change as follows:

- When \( x = 0 \), \( y = 0 \)
- When \( x = 1 \), \( y = 1 \)

Thus, we rewrite the first integral:

\[ \int_0^1 \frac{\left(\frac{y}{1-y}\right)^{m-1}}{(1+\frac{y}{1-y})^{m+n}} \cdot \frac{1}{(1-y)^2} \, dy \]

This simplifies to:

\[ \int_0^1 \frac{y^{m-1} (1-y)^{n-1}}{(1-y)^{m+n}} \, dy = I_{m,n} \]

### Step 3: Combine the Results

Now, we can express both integrals in terms of \( I_{m,n} \):

\[ \int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}} \, dx + \int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}} \, dx = I_{m,n} + I_{n,m} \]

Since \( I_{m,n} = I_{n,m} \), we have:

\[ 2 I_{m,n} \]

### Step 4: Relate to \( \alpha I_{m,n} \)

From the problem statement, we know that:

\[ \int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}} \, dx = \alpha I_{m,n} \]

Thus, we can equate:

\[ 2 I_{m,n} = \alpha I_{m,n} \]

### Step 5: Solve for \( \alpha \)

Assuming \( I_{m,n} \neq 0 \), we can divide both sides by \( I_{m,n} \):

\[ \alpha = 2 \]

### Final Answer

Thus, the value of \( \alpha \) is:

\[ \boxed{2} \]