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How do I integrate [math]\dfrac{1+x}{x^2 + 3x + 1} \mathrm{d}x[/math] ?
I haven't. Can you show me how to do that?
Sure. First, we need to factor the denominator. We can do that by completing the square: [math]x^2+3x+1 = (x+\frac{3}{2})^2 - \frac{1}{4} + 1[/math]. This simplifies to [math]\dfrac{1+x}{(x+\frac{3}{2})^2 + \frac{3}{4}}[/math].