8. $[7]$ Let $T=\int_{0}^{\ln 2} \frac{2 e^{3 x}+e^{2 x}-1}{e^{3 x}+e^{2 x}-e^{x}+1} d x$. Evaluate $e^{T}$.
Answer: $\frac{11}{4}$ Divide the top and bottom by $e^{x}$ to obtain that
$$
T=\int_{0}^{\ln 2} \frac{2 e^{2 x}+e^{x}-e^{-x}}{e^{2 x}+e^{x}-1+e^{-x}} d x
$$

Notice that $2 e^{2 x}+e^{x}-e^{-x}$ is the derivative of $e^{2 x}+e^{x}-1+e^{-x}$, and so
$$
T=\left[\ln \left|e^{2 x}+e^{x}-1+e^{-x}\right|\right]_{0}^{\ln 2}=\ln \left(4+2-1+\frac{1}{2}\right)-\ln 2=\ln \left(\frac{11}{4}\right)
$$

Therefore, $e^{T}=\frac{11}{4}$.