prove that $\int_{0}^{1}\displaystyle\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}dx=B(m,n)$
prove that $\int_{0}^{1}\displaystyle\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}dx=B(m,n)$
first note that the substitution $t \rightarrow 1-t$ gives us: $\int_{\frac{1}{2}}^1 t^{m-1}(1-t)^{n-1} \ dt = \int_0^{\frac{1}{2}} t^{n-1} (1-t)^{m-1} \ dt.$ therefore we have:
$B(m,n)=\int_0^1 t^{m-1}(1-t)^{n-1} \ dt = \int_0^{\frac{1}{2}}[ t^{m-1}(1-t)^{n-1} + t^{n-1}(1-t)^{m-1}] \ dt.$ now do the substitution: $t = \frac{x}{x+1}. \ \Box$