If f'(x) = frac{(sqrt{x}+1)e^{sqrt{x}}}{sqrt{ | vedclass

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(C) Step $1$: Integrate $f'(x)$ to find $f(x)$.$f(x) = \int \frac{(\sqrt{x}+1)e^{\sqrt{x}}}{\sqrt{x}} \, dx$Step $2$: Use substitution. Let $u = \sqrt

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$3e$

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(C) Step $1$: Integrate $f'(x)$ to find $f(x)$.$f(x) = \int \frac{(\sqrt{x}+1)e^{\sqrt{x}}}{\sqrt{x}} \, dx$Step $2$: Use substitution. Let $u = \sqrt{x}$, then $du = \frac{1}{2\sqrt{x}} \, dx$, which implies $dx = 2\sqrt{x} \, du = 2u \, du$.Step $3$: Substitute into the integral:$f(x) = \int \frac{(u+1)e^u}{u} \cdot 2u \, du = 2 \int (u+1)e^u \, du$Step $4$: Integrate by parts $\int (u+1)e^u \, du = (u+1)e^u - \int e^u \, du = (u+1)e^u - e^u + C = ue^u + C$.So, $f(x) = 2ue^u + C = 2\sqrt{x}e^{\sqrt{x}} + C$.Step $5$: Use $f(0) = e$ to find $C$.$f(0) = 2(0)e^0 + C = e \implies C = e$.Step $6$: Find $f(1)$.$f(1) = 2\sqrt{1}e^{\sqrt{1}} + e = 2e + e = 3e$.

If $f'(x) = \frac{(\sqrt{x}+1)e^{\sqrt{x}}}{\sqrt{x}}$ and $f(0) = e$, then $f(1) = \dots$

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