prove that $\int^1 _0 \frac{x^{m-1}(1-x)^{n-1}dx}{(a+x)^{m+n}} = \frac{\beta(m,n)}{a^n(a+1)^m}$ $$\int^1 _0 \frac{x^{m-1}(1-x)^{n-1}dx}{(a+x)^{m+n}} = \frac{\beta(m,n)}{a^n(a+1)^m}$$
While solving the above integral I tried making $a + x = t$ as one of the substitution but it turned out to be useless. I also tried substituting $(1-x)^{n-1} = t$ and $x^{m-1} = t$ but still could not prove the above result.
The substitution seems to be quite unconventional in this case, help me with this question.
 A: Try $u=(a+1)\frac{x}{a+x}$, then $\frac{1-x}{a+x}=\frac{1-u}a$ and $\mathrm{d}x=\frac{a(a+1)}{(a+1-u)^2}\,\mathrm{d}u$. Furthermore, $a+x=\frac{a(a+1)}{a+1-u}$. Thus,
$$
\begin{align}
\int_0^1\frac{x^{m-1}(1-x)^{n-1}\,\mathrm{d}x}{(a+x)^{m+n}}
&=\int_0^1\frac{u^{m-1}}{(a+1)^{m-1}}\frac{(1-u)^{n-1}}{a^{n-1}}\frac{\mathrm{d}u}{a(a+1)}\\
\end{align}
$$
A: This proof requires knowledge of the confluent hypergeometric function, 
${}_1F_1(a,b,x).$ It's not as easy as robjohn's. Please look at DLMF sec. 13 (Digital Library of Mathematical Functions) if you need more information. Incidentally, that reference uses M(a,b,x) for ${}_1F_1(a,b,x)$ and a bold-face M for the regularized version, so I'm sticking with ${}_1F_1$ to avoid confusion.
$$I(m,n,a):=\int_0^1 \frac{x^m (1-x)^{n-1}}{(a+x)^{m+n} } \,dx = \int_0^1 \int_0^\infty
\frac{1}{\Gamma(n+m)}\exp{\big(-(a+x)t\big)} t^{m+n-1} dt$$
by the Euler integral for the gamma function.  Interchange integrals, do the one in $x,$ and by integral representations implied in DLMF eq 13.4.1
$$I(m,n,a)=\frac{B(n,m)}{\Gamma(m+n)} \int_0^\infty e^{-a\,t} t^{m+n-1}
{}_1F_1(m,m+n,-t) \,dt$$ where $B(n,m)=\Gamma(m)\Gamma(n)/\Gamma(m+n).$
Use the Kummer transform DLMF 13.2.39 to get
$$I(m,n,a)=\frac{B(n,m)}{\Gamma(m+n)} \int_0^\infty e^{-(a+1)\,t} t^{m+n-1}
{}_1F_1(n,m+n,t) \,dt.$$  The point of this last step is to get a positive argument for the  ${}_1F_1$.  Use DLMF 13.10.4, which in   ${}_1F_1$ notation reads
$$ \int_0^\infty e^{-z\,t}\,t^{b-1} {}_1F_1(c,b,t) \, dt = 
z^{-b}(1-1/z)^{-c}\,\Gamma(b).$$  Letting $c=n,$ $b=n+m,$ $z=a+1,$ and algebra completes the proof.
A: Using the substitution
$$t=\frac{(1+p)\, u}{u+p}, \qquad dt=\frac{p \,(1+p)}{(p+u)^2} du,$$
the integral representation of the beta function
$$B(z,w)=\int_0^1 t^{z-1}(1-t)^{w-1}dt$$
gives
$$B(z,w)=\int_0^1 \left(\frac{(1+p)u}{(u+p)}\right)^{z-1}\left(1-\frac{(1+p)u}{(u+p)}\right)^{w-1}\left[\frac{p (1+p)}{(p+u)^2}\right]du.$$
Simplify,
$$B(z,w)=p^w(1+p)^z\int_0^1 \frac{u^{z-1}1-u)^{w-1}}{(u+p)^{z+w}}du$$
for $z>0,w>0$. Similarly, the substitution 
$$\frac{p}{u}-\frac{q}{t}=p-q$$
implies
$$B(z,w)=p^wq^z\int_0^1 \frac{u^{z-1}(1-u)^{w-1}}{(p+(q-p)u)^{z+w}}du,\qquad z>0,~~w>0.
$$
