Find $\int \dfrac{dx}{x^{2}+6x+10}$.

Solution The integrand contains the quadratic expression $x^{2}+6x+10$. So, we complete the square. $$x^{2}+6x+10=(x^{2}+6x+9)+1=(x+3)^{2}+1$$

Now we write the integral as $$\int \frac{dx}{x^{2}+6x+10}=\int \frac{dx}{(x+3)^{2}+1}$$

and use the substitution $u=x+3$. Then $du=dx$, and $$\int \frac{dx}{x^{2}+6x+10}=\int \frac{dx}{(x+3)^{2}+1}=\int \frac{du}{ u^{2}+1}=\tan ^{-1}u+C=\tan ^{-1}(x+3)+C$$