Conditions
integration from 0 to 1
of (m∧(m−1)((1−x)∧n−1)/((a+x)∧(m+n))
Solution
∫(a+x)m+nmm−1((1−x)n−1)dx=m+n−11mm−1(a+11−x)−n(a+x)−m−n+1((a+11−x)n−(1−x)2nF1(−m−n+1,−n;−m−n+2;a+1a+x)+constantm+n−1mm−1(a+x)−m−n+1−m+n−11mm−1(a+1)n(a+x)2−m−n+1F1(−m−n+1,−n;−m−n+2;a+1a+x)exp(−2iπn[2πarg(a+1)−2πarg(1−x)+21])+constant