Integrals Involving Quadratics — Question 6

Question 6 – Solution

Step 1: Write the possible perfect squares. Since the constant term is $9$, the square must be $$\begin{align*} (x+3)^2&=x^2+6x+9,\\ (x-3)^2&=x^2-6x+9. \end{align*}$$ Thus $$k=6\qquad\text{or}\qquad k=-6.$$ Step 2: Integrate when $k=6$. $$\begin{align*} \int\frac{dx}{x^2+6x+9} &=\int\frac{dx}{(x+3)^2}\\ &=\int(x+3)^{-2}\,dx\\ &=-(x+3)^{-1}+C=-\frac1{x+3}+C. \end{align*}$$ Step 3: Integrate when $k=-6$. $$\begin{align*} \int\frac{dx}{x^2-6x+9} &=\int(x-3)^{-2}\,dx\\ &=-(x-3)^{-1}+C=-\frac1{x-3}+C. \end{align*}$$ $$\boxed{k=6:\ -\frac1{x+3}+C;\qquad k=-6:\ -\frac1{x-3}+C}$$