Integrating Functions Using Long Division and Completing the Square AB & BC

A rational integrand is prepared before it is integrated, and which preparation depends on one comparison. If the numerator's degree is at least the denominator's, the fraction is improper and long division is required: $\frac{x^{2}}{x+1} = x - 1 + \frac{1}{x+1}$, whose three pieces are all routine while the original is not. Watch for the case where a common factor makes the division unnecessary, as in $\frac{x^{3}+x}{x^{2}+1} = x$.

For a proper fraction with an irreducible quadratic denominator, completing the square produces an arctangent: $x^{2}+4x+13 = (x+2)^{2}+9$ gives $\frac{1}{3}\arctan\!\left(\frac{x+2}{3}\right) + C$, where the $\frac{1}{3}$ comes from the chain rule and the shift has to appear inside the argument. If instead the numerator is the derivative of the denominator up to a constant, the answer is a logarithm and no completing is needed. Throughout, a numerator splits over a single denominator and a denominator never splits at all: $\frac{1}{x^{2}+1}$ is not $\frac{1}{x^{2}} + 1$, as $\frac{1}{2+3}$ against $\frac{1}{2}+\frac{1}{3}$ shows in one line.