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**Chris L T521** Let $\displaystyle \zeta=1+2x\implies x=\frac{1}{2}(\zeta-1)$

Thus, $\displaystyle \,d\zeta=2\,dx\implies\,dx=\frac{1}{2}\,d\zeta$

Limits change as well.. $\displaystyle \zeta(4)=9, \ \zeta{0}=1$

Thus, our integral is

$\displaystyle \frac{1}{4}\int_1^9 \frac{\zeta-1}{\sqrt{\zeta}}\,d\zeta=\frac{1}{4}\int_1^9 \bigg[\sqrt{\zeta}-(\zeta)^{-\frac{1}{2}}\bigg]$

$\displaystyle =\frac{1}{4}\left.\bigg[\frac{2}{3}\zeta^{\frac{3}{2}}-2\sqrt{\zeta}\bigg]\right|_1^9=\frac{1}{4}\bigg(12+\frac{4}{3}\bigg)= \color{red}\boxed{\frac{10}{3}}$