If l(m,n)=∫01 tm(1+t)ndt then the expression for l(m,n) in terms of l(m+1,n−1) is
2nm+1−nm+1l(m+1,n−1)
nm+1l(m+1,n−1)
2nm+1+nm+1l(m+1,n−1)
mn+1l(m+1,n−1)
I(m,n)=1m+1tm+1(1+t)n01
−nm+1I(m+1,n−1)⇒I(m,n)=2nm+1−nm+1I(m+1,n−1)