Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 13 / a / ii / Solution

Let $D=3x^2+2x+2$. Splitting the numerator into a derivative term and a constant gives $x-1=D^{\prime }/6-4/3$. ThusComplete the square as $D=((3x+1{)}^2+5)/3$ and substitute $u=3x+1$. The remaining integral is $\int du/(u^2+5)=\arctan (u/\sqrt{5})/\sqrt{5}$. HenceThe quadratic denominator is positive for all real $x$, so no logarithmic absolute value or real-domain exclusion is needed.