Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.

Using long division,
$$ \frac{x^2+4}{x^2+6x+10} = 1 -6\frac{x+1}{x^2+6x+10}= 1 -\frac{6x}{x^2+6x+10}-\frac{6}{x^2+6x+10}$$
Integrating $$1 -\frac{6x}{x^2+6x+10}-\frac{6}{x^2+6x+10} $$ should be straight forward
Note that $x^2+6x+10= (x+3)^2 +1$
