The average velocity for a turbulent pipe flow can be evaluated using the expression,

$u_{\text{a}\text{v}\text{g}}=\frac{Q}{A}=\frac{1}{R^2}\frac{{\displaystyle \underset{0}{\overset{R}{}}}}{b}ar(u)(r)dA=\frac{1}{R^2}\frac{{\displaystyle \underset{0}{\overset{R}{}}}}{b}ar(u)(r)2rdr=\frac{2}{R^2}\frac{{\displaystyle \underset{0}{\overset{R}{}}}}{b}ar(u)(y)(R-y)dy$

where the simple $y=R-r$ coordinate transformation is used, with $y$ being the dimenional distance

from the wall. Assuming that the law of the wall is accurate all the way to the wall, i$.$e$.,$ neglecting

the very thin viscous sublayer, substitute the log$-$law, $u^{+}={}^{-1}ln{y}^{+}+B,$ into the average velocity

definition to prove that

$\frac{u_{\text{a}\text{v}\text{g}}}{u^{**}}=\frac{1}{}ln(\frac{R{u}^{**}}{v})+B-\frac{3}{2}$

Before substituting $u^{+}={}^{-1}ln{y}^{+}+B$ into the average velocity integral, express it first as

$\frac{(\frac{?}{\text{b}\text{a}\text{r}}(u))(y)}{u^{**}}=\frac{1}{}ln(\frac{y{u}^{**}}{v})+B$