Split off a denominator derivative

The denominator derivative is $2x+4$. Rewrite $6x+16=3(2x+4)+4$. Hence split the integral into $3\int\frac{2x+4}{x^2+4x+13}\,dx+4\int\frac1{x^2+4x+13}\,dx$.

Complete the square: $x^2+4x+13=(x+2)^2+9$. The result is $3\ln|x^2+4x+13|+\frac43\tan^{-1}[(x+2)/3]+C=3\ln(x^2+4x+13)+\frac43\tan^{-1}[(x+2)/3]+C$. The modulus can be removed because $(x+2)^2+9>0$ for every real $x$, as the source explains.