If int frac{e^{sqrt{x}}}{sqrt{x}} (x + sqrt{x | vedclass

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(B) Let $I = \int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) dx$. Substitute $\sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{

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$2$

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(B) Let $I = \int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) dx$. Substitute $\sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{dx}{\sqrt{x}} = 2 dt$. Substituting these into the integral,we get: $I = \int e^t (t^2 + t) (2 dt) = 2 \int (t^2 + t) e^t dt$. Using the integration by parts formula $\int u v' dt = uv - \int u' v dt$,let $u = t^2 + t$ and $v' = e^t$. Then $u' = 2t + 1$ and $v = e^t$. $I = 2 [(t^2 + t) e^t - \int (2t + 1) e^t dt]$. Now,evaluate $\int (2t + 1) e^t dt$ using integration by parts again: Let $u = 2t + 1$ and $v' = e^t$. Then $u' = 2$ and $v = e^t$. $\int (2t + 1) e^t dt = (2t + 1) e^t - \int 2 e^t dt = (2t + 1) e^t - 2 e^t = (2t - 1) e^t$. Substituting this back into the expression for $I$: $I = 2 [ (t^2 + t) e^t - (2t - 1) e^t ] + K = 2 e^t [ t^2 + t - 2t + 1 ] + K = 2 e^t [ t^2 - t + 1 ] + K$. Since $t = \sqrt{x}$,we have $I = e^{\sqrt{x}} [ 2x - 2\sqrt{x} + 2 ] + K$. Comparing this with $e^{\sqrt{x}} [ Ax + B\sqrt{x} + C ] + K$,we get $A = 2$,$B = -2$,and $C = 2$. Therefore,$A + B + C = 2 - 2 + 2 = 2$.

If $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) dx = e^{\sqrt{x}} [Ax + B \sqrt{x} + C] + K$,then $A + B + C = $

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