Integrals Involving Quadratics — Question 3

Question 3 – Solution

Step 1: Rewrite the numerator using the derivative of the denominator. $$x=\frac12(2x+4)-2.$$ Step 2: Split the integral. $$\begin{align*} I={}&\frac12\int\frac{2x+4}{x^2+4x+8}\,dx -2\int\frac{dx}{x^2+4x+8}. \end{align*}$$ Step 3: Integrate the derivative term. $$\frac12\int\frac{2x+4}{x^2+4x+8}\,dx =\frac12\ln(x^2+4x+8).$$ Step 4: Complete the square. $$x^2+4x+8=(x+2)^2+4.$$ Therefore, $$\begin{align*} -2\int\frac{dx}{(x+2)^2+2^2} &=-2\left[\frac12\arctan\left(\frac{x+2}{2}\right)\right]\\ &=-\arctan\left(\frac{x+2}{2}\right). \end{align*}$$ Step 5: Combine the results. $$\boxed{\frac12\ln(x^2+4x+8)-\arctan\frac{x+2}{2}+C}$$