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Ta có $I=\int\limits_{0}^{\ln2}\frac{2e^{3x}+e^{2x}-1}{e^{3x}+e^{2x}-e^{x}+1} dx$$I=\int\limits_{0}^{\ln2}\frac{3e^{3x}+2e^{2x}-e^x-(e^{3x}+e^{2x}-e^{x}+1)}{e^{3x}+e^{2x}-e^{x}+1} dx$$I=\int\limits_{0}^{\ln2}\left[ {\frac{3e^{3x}+2e^{2x}-e^x}{e^{3x}+e^{2x}-e^{x}+1} -1} \right]dx$$I=\int\limits_{0}^{\ln2}\left[ {\frac{3e^{3x}+2e^{2x}-e^x}{e^{3x}+e^{2x}-e^{x}+1} } \right]dx -\int\limits_{0}^{\ln2}dx$$I=\int\limits_{0}^{\ln2}{\frac{d(e^{3x}+e^{2x}-e^{x}+1)}{e^{3x}+e^{2x}-e^{x}+1} } -\int\limits_{0}^{\ln2}dx$$I=\ln\left| {e^{3x}+e^{2x}-e^{x}+1} \right|_0^{\ln 2} - x |_0^{\ln 2}$ $\boxed{I=\ln\frac{11}{4}}$

Ta có $I=\int\limits_{0}^{\ln2}\frac{2e^{3x}+e^{2x}-1}{e^{3x}+e^{2x}-e^{x}+1} dx$$I=\int\limits_{0}^{\ln2}\frac{3e^{3x}+2e^{2x}-e^x-(e^{3x}+e^{2x}-e^{x}+1)}{e^{3x}+e^{2x}-e^{x}+1} dx$$I=\int\limits_{0}^{\ln2}\left[ {\frac{3e^{3x}+2e^{2x}-e^x}{e^{3x}+e^{2x}-e^{x}+1} -1} \right]dx$$I=\int\limits_{0}^{\ln2}\left[ {\frac{3e^{3x}+2e^{2x}-e^x}{e^{3x}+e^{2x}-e^{x}+1} } \right]dx -\int\limits_{0}^{\ln2}dx$$I=\int\limits_{0}^{\ln2}{\frac{d(e^{3x}+e^{2x}-e^{x}+1)}{e^{3x}+e^{2x}-e^{x}+1} } -\int\limits_{0}^{\ln2}dx$$I=\ln\left| {e^{3x}+e^{2x}-e^{x}+1} \right|_0^{\ln 2} - x |_0^{\ln 2}$ $\boxed{I=\ln\frac{11}{4}\Rightarrow e^I=\frac{11}{4}}$