If \(\int(2 x+4) \sqrt{x-1} \mathrm{~d} x=\mathrm{a}(x-1)^{5 / 2}+\mathrm{b}(x-1)^{3 / 2}+\mathrm{c}\) where c is a constant of integration, then the value of \((2 a+b)\) is

\(\begin{aligned}
& \text { Let } I=\int(2 x+4) \sqrt{x-1} d x \\
& \text { Let } x-1=\mathrm{t} \\
& \therefore \quad x=1+\mathrm{t} \\
& \therefore \quad \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad I=\int[2(1+t)+4] \sqrt{t} d t \\
& =\int\left(6 \sqrt{t}+2 t^{\frac{3}{2}}\right) d t \\
& =\frac{6 \mathrm{t}^{\frac{3}{2}}}{\frac{3}{2}}+2 \frac{\mathrm{t}^{\frac{5}{2}}}{\frac{5}{2}}+\mathrm{c} \\
& =4(x-1)^{\frac{3}{2}}+\frac{4}{5}(x-1)^{\frac{5}{2}}+\mathrm{c}
\end{aligned}\)
Comparing with \(\mathrm{a}(x-1)^{\frac{5}{2}}+\mathrm{b}(x-1)^{\frac{3}{2}}+\mathrm{c}\), we get
\(\begin{array}{cl}
& a=\frac{4}{5}, b=4 \\
\therefore \quad & 2 a+b=\frac{8}{5}+4=\frac{28}{5}
\end{array}\)