3. [4] Compute $e^{A}$ where $A$ is defined as
$$
\int_{3 / 4}^{4 / 3} \frac{2 x^{2}+x+1}{x^{3}+x^{2}+x+1} d x \text {. }
$$
Answer: $\frac{16}{9}$
Solution: We can use partial fractions to decompose the integrand to $\frac{1}{x+1}+\frac{x}{x^{2}+1}$, and then integrate the addends separately by substituting $u=x+1$ for the former and $u=x^{2}+1$ for latter, to obtain $\ln (x+1)+\left.\frac{1}{2} \ln \left(x^{2}+1\right)\right|_{3 / 4} ^{4 / 3}=\left.\ln \left((x+1) \sqrt{x^{2}+1}\right)\right|_{3 / 4} ^{4 / 3}=\ln \frac{16}{9}$. Thus $e^{A}=16 / 9$
Alternate solution: Substituting $u=1 / x$, we find
$$
A=\int_{4 / 3}^{3 / 4} \frac{2 u+u^{2}+u^{3}}{1+u+u^{2}+u^{3}}\left(-\frac{1}{u^{2}}\right) d u=\int_{3 / 4}^{4 / 3} \frac{2 / u+1+u}{1+u+u^{2}+u^{3}} d u
$$

Adding this to the original integral, we find
$$
2 A=\int_{3 / 4}^{4 / 3} \frac{2 / u+2+2 u+2 u^{2}}{1+u+u^{2}+u^{3}} d u=\int_{3 / 4}^{4 / 3} \frac{2}{u} d u
$$

Thus $A=\ln \frac{16}{9}$ and $e^{A}=\frac{16}{9}$.