int e^{sqrt{x}} , dx = . . . . . . + c ; x | vedclass

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Question

(A) To solve the integral $I = \int e^{\sqrt{x}} \, dx$,we use the substitution method. Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, d

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To solve the integral $I = \int e^{\sqrt{x}} \, dx$,we use the substitution method. Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$. Substituting these into the integral,we get: $I = \int e^t (2t) \, dt = 2 \int t e^t \, dt$. Using integration by parts,where $\int u \, dv = uv - \int v \, du$: Let $u = t$ and $dv = e^t \, dt$. Then $du = dt$ and $v = e^t$. $I = 2 [t e^t - \int e^t \, dt] = 2 [t e^t - e^t] + c = 2 e^t (t - 1) + c$. Substituting $t = \sqrt{x}$ back,we get: $I = 2 e^{\sqrt{x}} (\sqrt{x} - 1) + c$. Thus,the correct option is $A$.

$\int e^{\sqrt{x}} \, dx = $ . . . . . . $+ c ; x > 0$

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