Prove the following : $$ \int_0^{ 1 } \frac{x^{l-1} (1-x)^{m-1}}{ (b+cx)^{l+m} }\,dx = \frac{\beta(l,m)}{(b+c)^{l}b^m} $$
I found this substitution :
$$ x = \frac{-b}{t+c} $$
Using this, we get the integral as
$$ \dfrac{{-1}^{l-1}}{b^m} \int_{-\infty}^{-b-c} \dfrac{(t+b+c)^{m-1}}{t^{l+m}}dt $$
Then on further substituting $ t+b+b = y$ and then $ y=-(b+c)x $ I got the integral as
$$ \dfrac{1}{(b+c)^{l}b^m} \int_0^\infty \dfrac{x^{m-1}}{(1+x)^{l+m}}dx = \dfrac{\beta(l,m)}{(b+c)^{l}b^m} $$
Is there a better substitution to solve this problem?