Integration through partial fractions with complex roots In integrating the following: $\frac{1}{(x^2+2x+3)^2}$ I am trying to use partial fraction decomposition as follows:
$\frac{1}{(x^2+2x+3)^2} = \frac{Ax + B}{x^2+2x+3} + \frac{Cx + D}{(x^2+2x+3)^2}$
Which gives me:
$1 = (Ax + B)(x^2 +2x +3) + (Bx + C)$
And that gets me back to where I started.
$\frac{1}{(x^2+2x+3)^2}$
What am I doing wrong?

Hint:
Notice that $\left(x^2 + 2x + 3\right)^2\equiv\left((x + 1)^2 + 2\right)^2$. That is,
$$\int\frac1{\left(x^2 + 2x + 3\right)^2}\,\mathrm dx\equiv\int\frac1{\left((x + 1)^2 + 2\right)^2}\,\mathrm dx.$$
Let $u = x + 1\implies\mathrm du = \mathrm dx$. So,
$$\int\frac1{\left((x + 1)^2 + 2\right)^2}\,\mathrm dx\equiv\int\frac1{\left(u^2 + 2\right)^2}\,\mathrm du.$$
Now, apply an appropriate reduction formula.
