<s>[INST] An identity involving the Beta function I'm trying to show that 
$$ \int _0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx = \frac{B(a,b)}{(1+c)^ac^b}$$
Where $$B(a,b) :=  \int _0^1 x^{a-1}(1-x)^{b-1}dx $$ is the "Beta function".  I am supposed to use a substitution but I'm pretty much stuck.  I am familiar with the basic properties of the Beta function, its relation to the gamma function etc.     Any hints or advice you care to offer would be super cool. 
 [/INST] Let $\displaystyle   \int _0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx = I$
We have , 
$(1+c)^ac^bI=\displaystyle  \int _0^1 \frac{(1+c)^ac^bx^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx $
$=\displaystyle  \int _0^1 \frac{(1+c)c((1+c)x)^{a-1}(c(1-x))^{b-1}}{(x+c)^{a+b}}dx $
$=\displaystyle  \int _0^1 \left(\frac{(1+c)c}{(x+c)^2}\right)\left(\frac{(1+c)x}{(x+c)}\right)^{a-1}\left(\frac{c(1-x)}{(x+c)}\right)^{b-1}dx$
Let $\displaystyle y=\left(\frac{c(1-x)}{(x+c)}\right)$ and we have $\displaystyle dy=-\left(\frac{(1+c)c}{(x+c)^2}\right)dx$
Then we have the above $=-\displaystyle\int _1^0y^{b-1}(1-y)^{a-1}dy=\int_0^1y^{b-1}(1-y)^{a-1}dy=B(b,a)=B(a,b)$
So ultimately we have,
$\displaystyle(1+c)^ac^bI=B(a,b)\Rightarrow I=\frac{B(a,b)}{(1+c)^ac^b}$
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