int sqrt{x} e^{sqrt{x}} , dx = | vedclass

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(B) Let $I = \int \sqrt{x} e^{\sqrt{x}} \, dx$. Substitute $x = t^2$,so $dx = 2t \, dt$. Then $I = \int t e^t (2t \, dt) = 2 \int t^2 e^t \, dt$. Usin

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$(2x - 4\sqrt{x} + 4) e^{\sqrt{x}} + c$

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(B) Let $I = \int \sqrt{x} e^{\sqrt{x}} \, dx$. Substitute $x = t^2$,so $dx = 2t \, dt$. Then $I = \int t e^t (2t \, dt) = 2 \int t^2 e^t \, dt$. Using integration by parts $\int u v \, dt = u \int v \, dt - \int (u' \int v \, dt) \, dt$: Let $u = t^2$ and $v = e^t$. $I = 2 [t^2 e^t - \int 2t e^t \, dt] = 2t^2 e^t - 4 \int t e^t \, dt$. Applying integration by parts again for $\int t e^t \, dt$: $I = 2t^2 e^t - 4 [t e^t - \int e^t \, dt] = 2t^2 e^t - 4t e^t + 4e^t + c$. Substituting $t = \sqrt{x}$ back: $I = 2x e^{\sqrt{x}} - 4\sqrt{x} e^{\sqrt{x}} + 4e^{\sqrt{x}} + c = (2x - 4\sqrt{x} + 4) e^{\sqrt{x}} + c$.

$\int \sqrt{x} e^{\sqrt{x}} \, dx = $

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