Discussion
\(\displaystyle\int \frac{y^{2} - 3y + 2}{y^{2} + y}\, dy\) is equal to
Answer: B
The integral is \(y + 2\log|y| - 6\log|1 + y| + C\).
Divide first, then use partial fractions:
\[ \begin{aligned} \frac{y^{2} - 3y + 2}{y^{2} + y} &= 1 + \frac{2 - 4y}{y(y + 1)} \\ \frac{2 - 4y}{y(y + 1)} &= \frac{2}{y} - \frac{6}{y + 1} \end{aligned} \]
(At \(y = 0\) the numerator gives 2, and at \(y = -1\) it gives \(\frac{6}{-1} = -6\).) Integrating term by term gives \(y + 2\log|y| - 6\log|1 + y| + C\). The other options have a wrong coefficient or sign on one of the logarithms.
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