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 \[ \int \frac{dx}{x^{2}+6x+10}=\int \frac{dx}{(x+3)^{2}+1}=\int \frac{du}{ u^{2}+1}=\tan ^{-1}u+C=\tan ^{-1}(x+3)+C \]