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How do I integrate [math]\dfrac{1+x}{x^2 + 3x + 1} \mathrm{d}x[/math] ?
Okay, I see that. How does that help with partial fractions?
Now we can use partial fractions to write [math]\dfrac{1+x}{(x+\frac{3}{2})^2 + \frac{3}{4}}[/math] as [math]\dfrac{A}{x+\frac{3}{2}} + \dfrac{Bx+C}{(x+\frac{3}{2})^2 + \frac{3}{4}}[/math]. Solving for A, B, and C gives [math]A = \frac{3}{4}, B = -\frac{1}{2}, C = \frac{1}{4}[/math].