$I=\frac{\int\limits_{0}^{\ln2} }{\frac{22e^{3x}+e^{2x}-1}{e^{3x}+e^{2x}-e^x+1} } dx=\int\limits_{0}^{\ln2} \frac{3x^{3x}+2e^{2x}-e^x-(e^{3x}+e^{2x}-e^x+1)}{e^{3x}+e^{2x}-e^x+1} $$\int\limits_{0}^{\ln2}(\frac{3e^{3x}+2e^{2x}-e^x}{e^{3x}+e^{2x}-e^x+1} )dx $$\ln(e^{3x}+e^{2x}-e^x+1 )\left| \begin{gathered} \ln2\\ 0 \\ \end{gathered} \right..-x\left| \begin{gathered} \ln2 \\ 0 \\ \end{gathered} \right.=\ln11-\ln4=\ln\frac{11}{4} $Vậy $e^I=\frac{11}{4} $
$I=\frac{\int\limits_{0}^{\ln2} }{\frac{22e^{3x}+e^{2x}-1}{e^{3x}+e^{2x}-e^x+1} } dx=\int\limits_{0}^{\ln2} \frac{3x^{3x}+2e^{2x}-e^x-(e^{3x}+e^{2x}-e^x+1)}{e^{3x}+e^{2x}-e^x+1} $$\int\limits_{0}^{\ln2}(\frac{3e^{3x}+2e^{2x}-e^x}{e^{3x}+e^{2x}-e^x+1} )dx $$\ln(e^{3x}+e^{2x}-e^x+1 )\left| \begin{gathered} \ln2\\ 0 \\ \end{gathered} \right..-x\left| \begin{gathered} \ln2 \\ 0 \\ \end{gathered} \right.=\ln11-\ln4=\ln\frac{11}{4} $Vậy $e^I=\frac{11}{4} $
$I=\frac{\int\limits_{0}^{\ln2} }{\frac{22e^{3x}+e^{2x}-1}{e^{3x}+e^{2x}-e^x+1} } dx=\int\limits_{0}^{\ln2} \frac{3x^{3x}+2e^{2x}-e^x-(e^{3x}+e^{2x}-e^x+1)}{e^{3x}+e^{2x}-e^x+1} $$\int\limits_{0}^{\ln2}(\frac{3e^{3x}+2e^{2x}-e^x}{e^{3x}+e^{2x}-e^x+1} )dx $$\ln(e^{3x}+e^{2x}-e^x+1 )\left| \begin{gathered} \ln2\\ 0 \\ \end{gathered} \right..-x\left| \begin{gathered} \ln2 \\ 0 \\ \end{gathered} \right.=\ln11-\ln4=\ln\frac{11}{4} $Vậy $e^I=\frac{11}{4} $