If ∫ e x x x + x d x = e x A x + B x + C + K , then A + B + C =

$I=\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\left(x+\sqrt{x}\right)dx$

Let $\sqrt{x}=t$

$\frac{1}{2\sqrt{x}}dx=dt$

i.e. $I=2\int e^t\left(t^2+t\right)dt$

$=2\int t^2{e}^{t}dt+2\int t{e}^{t}dt$

$=2t^2{e}^t-2\int 2t{e}^{t}dt+2\int t{e}^{t}dt$

$=2t^2{e}^t-2\int t{e}^{t}dt$

$=2t^2{e}^t-2\left[t{e}^t-\int e^{t}dt\right]=2t^2{e}^t-2t{e}^t+2e^t+C^{\text{'}}$

$=e^{\sqrt{x}}\left(2x-2\sqrt{x}+2\right)+C^{\text{'}}$

Hence, $A+B+C=2$