Evaluate the following integral by using beta and gamma functions $\int_{0}^{1}\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}.}$ How to evaluate this Evaluate the following integral by using beta and gamma functions $\int_{0}^{1}\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}.}$ 
How to evaluate this?  
I took $\frac{2x}{1+x}=y$ and trying to convert into beta function, but I could not get the answer.
 A: $$\int\limits_{0}^{1} \frac{x^{m}+x^{n}}{(1+x)^{m+n}} \mathrm{d}x = \mathrm{B}(m,n)$$
Proof:
We begin with the basic integral definition of the beta function
\begin{equation}
\mathrm{B}(x,y) = \int\limits^{1}_{0} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\label{eq:bf1}
\tag{1}
\end{equation}
for $\Re x \gt 0$ and $\Re y \gt 0$.
\begin{equation}
\mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v
\label{eq:bf2}
\tag{2}
\end{equation}
To derive this equation, we begin with \eqref{eq:bf1} and make the substitution
\begin{equation}
t = \frac{v}{1+v}
\end{equation}
to obtain
\begin{equation}
\mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{v^{x-1}}{(1+v)^{x-1}} \Big(1 - \frac{v}{1+v}\Big)^{y-1} \frac{1}{(1+v)^{2}} \mathrm{d} v
\end{equation}
simplification yields \eqref{eq:bf2}.
\begin{equation}
\mathrm{B}(x,y) = \int\limits_{0}^{1} \frac{(v^{x-1}+v^{y-1})}{(1+v)^{x+y}} \mathrm{d} v
\label{eq:bf3}
\tag{3}
\end{equation}
To obtain \eqref{eq:bf3} we begin with \eqref{eq:bf2} and break up the integral
\begin{equation}
\mathrm{B}(x,y) = \int\limits_{0}^{1} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v + \int\limits_{1}^{\infty} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v
\end{equation}
and designate the last integral as I.
For I, we make the substitution $w = v^{-1}$
\begin{equation}
\mathrm{I} = \int\limits_{0}^{1} \frac{1}{w^{x-1}} \frac{w^{x+y}}{(1+w)^{x+y}} \frac{1}{w^{2}} \mathrm{d} w
\end{equation}
Simplifying and then making the substitution for I yields \eqref{eq:bf3}.
Note that equation \eqref{eq:bf3} for the beta function appears as Equation 3 on page 9 (pdf page 35) of Higher Transcendental Functions (Bateman Manuscript), Volume 1.
A: Expanding Nemo's answer,
setting
$x = \frac1{y}$,
so
$dy = -\frac{dy}{y^2}$.
Then
$\begin{array}\\
\int_{0}^{1}\frac{x^{n-1}}{(1+x)^{m+n}}dx
&=-\int_{\infty}^{1}\frac{1}{y^{n-1}(1+1/y)^{m+n}}\frac{dy}{y^2}\\
&=\int_1^{\infty}\frac{y^{m+n}dy}{y^{n+1}(y+1)^{m+n}}\\
&=\int_1^{\infty}\frac{y^{m-1}dy}{(y+1)^{m+n}}\\
\end{array}
$
and his result is verified.
Note:
Here is Nem's answer which,
for some reason,
was deleted:
\begin{align}
\int_{0}^{1}\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}dx&=\int_{0}^{1}\frac{x^{m-1}}{(1+x)^{m+n}}dx+\int_{0}^{1}\frac{x^{n-1}}{(1+x)^{m+n}}dx\\
&=\int_{0}^{1}\frac{x^{m-1}}{(1+x)^{m+n}}dx+\int_{1}^{\infty}\frac{x^{m-1}}{(1+x)^{m+n}}dx\\
&=\int_{0}^{\infty}\frac{x^{m-1}}{(1+x)^{m+n}}dx\\
&=B(m,n)
\end{align}
