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I'm taking an undergraduate statistics course. I have a problem in which $X\sim BETA(a,b)$ and $Y\sim BETA(a+b,c)$,first I found the joint PDF of $U=XY$ and $V=X$ which is the function I'm integrating. I'm trying to show that $U\sim BETA(a,b+c)$. The solution has a step where the integral to be evaluated transforms from this,

\begin{equation*} \int^1_u\frac{\Gamma(a+b+c)}{\Gamma(a)\Gamma(b)\Gamma(c)}v^{-(b+c)}(1-v)^{b-1}u^{a+b-1}(v-u)^{c-1}dv \end{equation*}

to this.
\begin{equation*} =\frac{\Gamma(a+b+c)}{\Gamma(a)\Gamma(b+c)}u^{a-1}(1-u)^{b+c-1}\int^1_u\frac{\Gamma(b+c)}{\Gamma(b)\Gamma(c)}\frac{v^{-(b+c)}(1-v)^{b-1}u^{b-1}(v-u)^{c-1}}{(1-u)^{b+c-1}}dv \end{equation*}

The leftover integral should evaluate to $1$. Then the left hand side is the answer sought. I'm having trouble understanding how the function, especially the fraction with individual gamma functions, simplifies or breaks apart. I thought maybe it has to do with the original $X$ and $Y$ distributions. But there is no $(b+c)$ term involved.

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    for starters: pull the $v$-indpendent rubbish out of the integral – 2017-02-13
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    I wholeheartedly second @Tired here. Half of the stuff in your integral is irrelevant. Pull it outside the integral. – 2017-02-13

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