Discussion
\(\displaystyle \int \frac{x^{2} + 6x + 1}{(x + 3)^{2}}\,dx =\)
Answer: C
The integral is \(x + \frac{8}{x + 3} + C\).
Rewrite the numerator around \((x + 3)^{2} = x^{2} + 6x + 9\):
\[ \begin{aligned} \frac{x^{2} + 6x + 1}{(x + 3)^{2}} &= \frac{(x + 3)^{2} - 8}{(x + 3)^{2}} \\ &= 1 - \frac{8}{(x + 3)^{2}} \end{aligned} \]
\[ \begin{aligned} &\int\left(1 - \frac{8}{(x + 3)^{2}}\right)dx \\ &= x + \frac{8}{x + 3} + C \end{aligned} \]
Because \(\int (x + 3)^{-2}dx = -(x + 3)^{-1}\), the minus signs combine to give \(+\frac{8}{x + 3}\). The option \(x - \frac{8}{x + 3}\) misses this sign change.
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