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

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Question

(A) 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 $\fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) 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 $\frac{dx}{\sqrt{x}} = 2dt$. Substituting these into the integral: $I = \int e^t (t^2 + t) (2 dt) = 2 \int e^t (t^2 + t) \, dt$. Using the formula $\int e^t [f(t) + f'(t)] \, dt = e^t f(t) + C$,let $f(t) = t^2$. Then $f'(t) = 2t$. This does not fit directly. Let's use integration by parts or the method of undetermined coefficients. $2 \int e^t (t^2 + t) \, dt = 2 [e^t t^2 - \int e^t (2t) \, dt + \int e^t t \, dt] = 2 [e^t t^2 - \int e^t t \, dt]$. Alternatively,$2 \int e^t (t^2 + t) \, dt = 2 e^t (t^2 - t + 1) + C$. Substituting $t = \sqrt{x}$ back: $I = 2 e^{\sqrt{x}} (x - \sqrt{x} + 1) + C$.

$\int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) \, dx$

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