If the value of the integral int_{1}^{2} e^{x | vedclass

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(B) Let $I = \int_{e}^{e^4} \sqrt{\ln x} dx$. Substitute $\ln x = t^2$,which implies $x = e^{t^2}$. Then $dx = 2t e^{t^2} dt$. When $x = e$,$t^2 = \ln

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$2e^4 - e - \alpha$

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(B) Let $I = \int_{e}^{e^4} \sqrt{\ln x} dx$. Substitute $\ln x = t^2$,which implies $x = e^{t^2}$. Then $dx = 2t e^{t^2} dt$. When $x = e$,$t^2 = \ln e = 1$,so $t = 1$. When $x = e^4$,$t^2 = \ln e^4 = 4$,so $t = 2$. Thus,$I = \int_{1}^{2} t (2t e^{t^2}) dt = \int_{1}^{2} 2t^2 e^{t^2} dt$. Using integration by parts,let $u = t$ and $dv = 2t e^{t^2} dt$. Then $du = dt$ and $v = e^{t^2}$. Applying the formula $\int u dv = uv - \int v du$: $I = [t e^{t^2}]_{1}^{2} - \int_{1}^{2} e^{t^2} dt$. $I = (2 e^{2^2} - 1 e^{1^2}) - \alpha$. $I = 2e^4 - e - \alpha$.

If the value of the integral $\int_{1}^{2} e^{x^2} dx$ is $\alpha$,then the value of $\int_{e}^{e^4} \sqrt{\ln x} dx$ is:

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