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One of the points of intersection of the curves $y=1+3 x-2 x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(l \sqrt{5}+\mathrm{m})-\mathrm{n} \log _{\mathrm{e}}(1+\sqrt{5})$, where $l, \mathrm{~m}, \mathrm{n} \in \mathbf{N}$. Then $l+\mathrm{m}+\mathrm{n}$ is equal to
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$\begin{aligned} & A=\int_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}}\left(1+3 x-2 x^2-\frac{1}{x}\right) \mathrm{dx} \\ & A=\left[x+\frac{3 x^2}{2}-\frac{2 x^3}{3}-\ln x\right]_{\frac{1}{2}}^{1+\sqrt{5}} \\ & A=\frac{1+\sqrt{5}}{2}+\frac{3}{2}\left(\frac{1+\sqrt{5}}{2}\right)^2-\frac{2}{3}\left(\frac{1+\sqrt{5}}{2}\right)^3-\ln \left(\frac{1+\sqrt{5}}{2}\right) \\ & -\frac{1}{2}-\frac{3}{2}\left(\frac{1}{4}\right)+\frac{2}{3}\left(\frac{1}{8}\right)+\ln \left(\frac{1}{2}\right) \\ & A=\frac{1}{2}+\frac{\sqrt{5}}{2}+\frac{3}{8}+\frac{3}{4} \sqrt{5}+\frac{15}{8}-\frac{4}{3}-\frac{2}{3} \sqrt{5} \\ & -\frac{1}{2}-\frac{3}{8}+\frac{1}{12}-\ln (1+\sqrt{5}) \\ & =\sqrt{5}\left(\frac{1}{2}+\frac{3}{4}-\frac{2}{3}\right)+\frac{15}{8}-\frac{4}{3}+\frac{1}{12}-\ln (1+\sqrt{5}) \\ & =\frac{14}{24} \sqrt{5}+\frac{15}{24}-\ln (1+\sqrt{5})\end{aligned}$
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