integrate (x-3)/(x^2 + 2x+4).
note that $\displaystyle \int \frac {x - 3}{x^2 + 2x + 4}~dx = \int \frac {x + 1}{x^2 + 2x + 4}~dx - \int \frac 4{x^2 + 2x + 3}$
for the first integral, a substitution of $\displaystyle u = x^2 + 2x + 4$ works.
for the second integral, note that $\displaystyle \frac 4{x^2 + 2x + 4} = \frac 4{(x + 1)^2 + 3} = \frac 43 \cdot \frac 1{\left( \frac {x + 1}{\sqrt{3}} \right)^2 + 1}$
write the integral in that form and then do a substitution of $\displaystyle u = \frac {x + 1}{\sqrt{3}}$. you should be able to handle it from there