$\underset{n}{lim}\frac{5}{n}{\displaystyle \underset{i=0}{\overset{n}{}}}ln(1+\frac{5i}{n})=$

${\displaystyle \underset{2}{\overset{7}{}}}ln(x+1)dx$

${\displaystyle \underset{2}{\overset{7}{}}}lnxdx$

${\displaystyle \underset{0}{\overset{5}{}}}ln(x-1)dx$

${\displaystyle \underset{2}{\overset{7}{}}}ln(x-1)dx$