33. [18] Compute
$$
\int_{1}^{2} \frac{9 x+4}{x^{5}+3 x^{2}+x} d x .
$$
(No, your TI-89 doesn't know how to do this one. Yes, the end is near.)
Answer: $\ln \frac{80}{23}$. We break the given integral into two pieces:
$$
\int_{1}^{2} \frac{9 x+4}{x^{5}+3 x^{2}+x} d x=5 \int_{1}^{2} \frac{x^{4}+3 x+1}{x^{5}+3 x^{2}+x} d x-\int_{1}^{2} \frac{5 x^{4}+6 x+1}{x^{5}+3 x^{2}+x} d x .
$$

These two new integrals are easily computed; for, the first integrand reduces to $1 / x$ and the second is of the form $f^{\prime}(x) / f(x)$. We obtain
$$
\left[5 \ln |x|-\ln \left|x^{5}+3 x^{2}+x\right|\right]_{1}^{2}=\ln 32-\ln 46+\ln 5=\ln \frac{80}{23}
$$

Motivation. Writing $f(x)=9 x+4$ and $g(x)=x^{5}+3 x^{2}+x=x\left(x^{4}+3 x+1\right)$, we wish to find the antiderivative of $f(x) / g(x)$. It makes sense to consider other rational functions with denominator $g(x)$ that have an exact antiderivative. Clearly, if the numerator were $f_{1}(x)=x^{4}+3 x+1$ or a constant multiple, then we can integrate the function. Another trivial case is if the numerator were $f_{2}(x)=g^{\prime}(x)=5 x^{4}+6 x+1$ or a constant multiple. Guessing that $f(x)$ is a linear combination of $f_{1}(x)$ and $f_{2}(x)$, we easily find that $f(x)=9 x+4=5 f_{1}(x)-f_{2}(x)$.