If `I(m,n)=int_0^1x^(m-1)(1-x)^(n-1)dx`, then

Options

A. `I(m,n)=int_(0)^(oo) (x^(m-1))/((1+x)^(m-n))dx`
B. `I(m,n)=int_(0)^(oo) (x^(m))/((1+x)^(m+n))dx`
C. `I(m,n)=int_(0)^(oo) (x^(n-1))/((1+x)^(m+n))dx`
D. `I(m,n)=int_(0)^(oo)(x^(n))/((1+x)^(m+n))dx`

Explanation

To solve the integral \( I(m,n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx \), we will use a substitution method to transform the integral into a more manageable form.

### Step-by-Step Solution:

1. **Substitution**: Let \( x = \frac{1}{1+y} \). Then, we need to find \( dx \) in terms of \( dy \).
\[ dx = -\frac{1}{(1+y)^2} \, dy \]
When \( x = 0 \), \( y \to \infty \) and when \( x = 1 \), \( y = 0 \).

2. **Change of Limits**: The integral changes from:
\[ I(m,n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx \quad \text{to} \quad I(m,n) = \int_\infty^0 \left( \frac{1}{1+y} \right)^{m-1} \left( 1 - \frac{1}{1+y} \right)^{n-1} \left(-\frac{1}{(1+y)^2}\right) dy \]

3. **Simplifying the Integral**: The expression \( 1 - \frac{1}{1+y} = \frac{y}{1+y} \) leads to:
\[ I(m,n) = \int_\infty^0 \left( \frac{1}{1+y} \right)^{m-1} \left( \frac{y}{1+y} \right)^{n-1} \left(-\frac{1}{(1+y)^2}\right) dy \]
This simplifies to:
\[ I(m,n) = \int_0^\infty \frac{y^{n-1}}{(1+y)^{m+n}} \, dy \]

4. **Final Form**: The integral now has the form:
\[ I(m,n) = \int_0^\infty \frac{y^{n-1}}{(1+y)^{m+n}} \, dy \]

5. **Conclusion**: Thus, we find that:
\[ I(m,n) = \int_0^\infty \frac{x^{n-1}}{(1+x)^{m+n}} \, dx \]

### Final Answer:
The correct expression for \( I(m,n) \) is:
\[ I(m,n) = \int_0^\infty \frac{x^{n-1}}{(1+x)^{m+n}} \, dx \]