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

\(\begin{array}{ll} & \text { Let } I=\int(2 x+4) \sqrt{x-1} d x \\ & \text { Put } x-1=t^2 \\ & d x=2 t d t \\ \therefore \quad & I=\int\left(2\left(t^2+1\right)+4\right) \cdot t \times 2 t d t\end{array}\)
\(\begin{aligned}
& =\int\left(2 \mathrm{t}^2+2+4\right) 2 \mathrm{t}^2 \cdot \mathrm{dt} \\
& =\int\left(2 \mathrm{t}^2+6\right) 2 \mathrm{t}^2 \mathrm{dt} \\
& =\int\left(4 \mathrm{t}^4+12 \mathrm{t}^2\right) \mathrm{dt} \\
& =4 \int \mathrm{t}^4 \mathrm{dt}+12 \int \mathrm{t}^2 \mathrm{dt} \\
& =4 \frac{\mathrm{t}^5}{5}+12 \cdot \frac{\mathrm{t}^3}{3}+\mathrm{C} \\
& =\frac{4}{5} \mathrm{t}^5+4 \mathrm{t}^3+\mathrm{C} \\
& =\frac{4}{5}(x-1)^{\frac{5}{2}}+4(x-1)^{\frac{3}{2}}+C
\end{aligned}\)
But \(\int(2 x+4) \sqrt{x-1} \mathrm{~d} x=\mathrm{a}(x-1)^{\frac{5}{2}}+\mathrm{b}(x-1)^{\frac{3}{2}}+\mathrm{C}\)
\(\begin{aligned} & \therefore \quad a=\frac{4}{5} \text { and } b=4 \\ & \therefore \quad a+b=\frac{4}{5}+4=\frac{24}{5}\end{aligned}\)