Evaluate dx1x2+2x+9\displaystyle \int\frac{{\left.{d}{x}\right.}}{{{1}{x}^{{2}}+{2}{x}+{9}}}.
1x2+2x+9\displaystyle {1}{x}^{{2}}+{2}{x}+{9} is an irreducible quadratic since the value of b24ac\displaystyle {b}^{{2}}-{4}{a}{c} =   is negative.
Transform 1x2+2x+9\displaystyle {1}{x}^{{2}}+{2}{x}+{9} to d(e(x+f)2+1)\displaystyle {d}{\left({e}{\left({x}+{f}\right)}^{{2}}+{1}\right)} for constants d, e and f by completing the square and doing algebra.
d =       e =       f =  
Now make the substitution u=e(x+f)\displaystyle {u}=\sqrt{{{e}}}{\left({x}+{f}\right)} to obtain an integral of the form gduu2+1\displaystyle {g}\int\frac{{{d}{u}}}{{{u}^{{2}}+{1}}} for a constant g.
Finally integrate.
dx1x2+2x+9\displaystyle \int\frac{{\left.{d}{x}\right.}}{{{1}{x}^{{2}}+{2}{x}+{9}}} =