Integrals Involving Quadratics — Question 1

Question 1 – Solution

Step 1: Complete the square. $$\begin{align*} x^2+6x+13 &=(x^2+6x+9)+4\\ &=(x+3)^2+4. \end{align*}$$ Step 2: Rewrite the integral. $$I=\int\frac{dx}{(x+3)^2+2^2}.$$ Step 3: Use the arctangent formula. Let $u=x+3$, so $du=dx$: $$\begin{align*} I&=\int\frac{du}{u^2+2^2}\\ &=\frac12\arctan\left(\frac{u}{2}\right)+C\\ &=\frac12\arctan\left(\frac{x+3}{2}\right)+C. \end{align*}$$ $$\boxed{\frac12\arctan\frac{x+3}{2}+C}$$