Discussion
\(\int \dfrac{2x^{2} + 4x + 3}{x^{2} + x + 1}\,dx =\)
Answer: D
The integral is \(2x + log_{e}|x^{2} + x + 1| + C\).
Divide first: \(2x^{2} + 4x + 3 = 2(x^{2} + x + 1) + (2x + 1)\), so
\[ \begin{aligned} &\frac{2x^{2} + 4x + 3}{x^{2} + x + 1} \\ &= 2 + \frac{2x + 1}{x^{2} + x + 1} \end{aligned} \]
The second fraction has its numerator equal to the derivative of its denominator, so it integrates to \(\log_{e}|x^{2} + x + 1|\), and the 2 integrates to \(2x\). Check by differentiating: \(2 + \frac{2x + 1}{x^{2} + x + 1}\). Option E swaps the coefficients (\(x + 2\log\)), and options A to C omit the polynomial part.
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