'An' Integral

$\int \frac{(x-1)^3}{\sqrt{x}}dx = \int \frac{(x-1)^3}{\sqrt{x}}dx = \int \frac{x^3-3x^2+3x-1}{\{x^{1/2}}dx$

$\Rightarrow \int \frac{(x-1)^3}{\sqrt{x}}dx = \int (x^{5/2}-3x^{3/2}+3x^{1/2}-x^{-1/2})dx = \frac{2}{7}x^{7/2}-\frac{6}{5}x^{5/2}+2x^{3/2}-2x^{1/2}$

$\Rightarrow a=2, b=7, d=6, e=5, u=2, v=-2 \Rightarrow a+b+d+e+u+v = 2+7+6+5+2-2 = 20$

$\begin{aligned} I & = \int \frac {(x-1)^3}{\sqrt x}dx & \small \color{#3D99F6} \text{Let }u^2 = x \implies 2u\ du = dx \\ & = \int \frac {(u^2-1)^3}u \cdot 2u\ du \\ & = 2 \int \left(u^6-3u^4+3u^2-1\right)\ du \\ & = \frac 27u^7 - \frac 65 u^5 + 2u^3 - 2u + \color{#3D99F6}C & \small \color{#3D99F6} \text{where } C \text{ is the constant of integration.} \\ & = \frac 27x^\frac 72 - \frac 65 x^\frac 52 + 2x^\frac 32 - 2x^\frac 12 + C & \small \color{#3D99F6} \text{Putting back } u = \sqrt x \end{aligned}$

Therefore, $a+b+d+e+u+v = 2+7+6+5+2-2 = \boxed{20}$ .