To solve the given integral
\[
I = \int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}} \, dx,
\]
we will break it down into manageable steps.
### Step 1: Split the Integral
We can separate the integral into two parts:
\[
I = \int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}} \, dx + \int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}} \, dx.
\]
Let's denote these two integrals as \( I_1 \) and \( I_2 \):
\[
I_1 = \int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}} \, dx,
\]
\[
I_2 = \int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}} \, dx.
\]
### Step 2: Evaluate \( I_2 \) using Substitution
For \( I_2 \), we will use the substitution \( x = \frac{1}{y} \). Then, the limits change as follows:
- When \( x = 0 \), \( y = \infty \).
- When \( x = 1 \), \( y = 1 \).
The differential \( dx \) becomes \( dx = -\frac{1}{y^2} \, dy \). Thus, we have:
\[
I_2 = \int_\infty^1 \frac{\left(\frac{1}{y}\right)^{n-1}}{(1+\frac{1}{y})^{m+n}} \left(-\frac{1}{y^2}\right) \, dy.
\]
Rearranging gives:
\[
I_2 = \int_1^\infty \frac{y^{-(n-1)}}{\left(\frac{y+1}{y}\right)^{m+n}} \cdot \frac{1}{y^2} \, dy = \int_1^\infty \frac{y^{-(n-1)}}{\left(\frac{y+1}{y}\right)^{m+n}} \cdot \frac{1}{y^2} \, dy.
\]
This simplifies to:
\[
I_2 = \int_1^\infty \frac{y^{-(n+1)}}{(y+1)^{m+n}} \, dy.
\]
### Step 3: Combine \( I_1 \) and \( I_2 \)
Now we can combine \( I_1 \) and \( I_2 \):
\[
I = I_1 + I_2 = \int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}} \, dx + \int_1^\infty \frac{y^{-(n+1)}}{(y+1)^{m+n}} \, dy.
\]
### Step 4: Change of Variable in \( I_2 \)
Notice that \( I_2 \) can be rewritten with a change of variable \( y = x \):
\[
I_2 = \int_0^\infty \frac{x^{n-1}}{(1+x)^{m+n}} \, dx.
\]
### Step 5: Final Expression
Thus, we have:
\[
I = \int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}} \, dx + \int_0^\infty \frac{x^{n-1}}{(1+x)^{m+n}} \, dx.
\]
Using the symmetry property \( g(m, n) = g(n, m) \):
\[
I = g(m, n).
\]
### Conclusion
The final result is:
\[
I = g(m, n).
\]
