int e^{sqrt{x}} , dx is equal to (A is an arb | vedclass

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Question

(C) Let $I = \int e^{\sqrt{x}} \, dx$. Substitute $\sqrt{x} = t$,so $x = t^2$ and $dx = 2t \, dt$. Substituting these into the integral: $I = \int e^t

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$2(\sqrt{x} - 1) e^{\sqrt{x}} + A$

Vedclass · Answer

(C) Let $I = \int e^{\sqrt{x}} \, dx$. Substitute $\sqrt{x} = t$,so $x = t^2$ and $dx = 2t \, dt$. Substituting these into the integral: $I = \int e^t \cdot 2t \, dt = 2 \int t e^t \, dt$. Using integration by parts,where $\int u \, dv = uv - \int v \, du$: Let $u = t$ and $dv = e^t \, dt$,then $du = dt$ and $v = e^t$. $I = 2 \left( t e^t - \int e^t \, dt \right) + A = 2(t e^t - e^t) + A = 2(t - 1) e^t + A$. Substituting back $t = \sqrt{x}$: $I = 2(\sqrt{x} - 1) e^{\sqrt{x}} + A$.

$\int e^{\sqrt{x}} \, dx$ is equal to ($A$ is an arbitrary constant).

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