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How do I integrate [math]\dfrac{1+x}{x^2 + 3x + 1} \mathrm{d}x[/math] ?
Okay, I understand. Now what?
We can rewrite the integral as [math]\int \dfrac{\frac{3}{4}}{x+\frac{3}{2}} - \dfrac{\frac{1}{2}x+\frac{1}{4}}{(x+\frac{3}{2})^2 + \frac{3}{4}} \mathrm{d}x[/math]. The first term is easy to integrate, while the second term can be tackled using substitution.