Integrals Involving Quadratics

Completing the square is simply rewriting a quadratic like $x^2 + 4x + 13$ as a perfect square plus a leftover number, $(x + 2)^2 + 9$. The name comes from geometry: picture $x^2 + 4x$ as a square plus two strips, and you add the small corner piece that completes a bigger square. It's good for integrating fractions and roots with quadratics that won't factor, and for finding the peak of a circuit's response or the center of a bell curve. All we're really saying with $\int \frac{dx}{(x - h)^2 + k^2} = \frac1k\arctan\frac{x - h}{k} + C$ is: $h$ (how far the parabola is slid sideways, where its lowest point sits) and $k^2$ (how far above zero that lowest point is) turn the integral into $\int \frac{dx}{1 + x^2}$'s arctangent, just shifted and stretched. It looks heavy, but an old algebra trick is doing all the work; bear with it.

Look at $\int \frac{dx}{x^2 + 4x + 13}$. You can't use partial fractions, because $x^2 + 4x + 13$ has no real roots ($16 - 52 < 0$). But you do know $\int \frac{du}{u^2 + 9} = \frac13\arctan\frac u3$. If only the denominator were a square plus a number... and it is, once you complete the square.

Completing the square:

For example, $x^2 + 4x + 13 = (x + 2)^2 + 9$. Now substitute $u = x + 2$, so $du = dx$, and it's the arctangent integral exactly. The shift costs nothing, which is why this works so cleanly.

A linear numerator. For $\displaystyle\int \frac{px + q}{x^2 + bx + c}\,dx$, there's one more step. Remember the log rule from substitution: when the top is the derivative of the bottom, you get a log,

So write the numerator as a multiple of the denominator's derivative, $2x + b$, plus whatever constant is left over. The derivative part gives the logarithm, and the leftover constant gives an arctangent after you complete the square.

Inside a root. Completing the square also sets up trig substitution. $\sqrt{-x^2 + 6x - 5} = \sqrt{4 - (x - 3)^2}$ is ready for $x - 3 = 2\sin\theta$, and $\sqrt{x^2 + 2x + 5} = \sqrt{(x + 1)^2 + 4}$ is ready for $x + 1 = 2\tan\theta$.

A common mistake. With a leading minus sign, as in $-x^2 + 6x - 5$, it's tempting to complete the square as if the minus weren't there. Factor it out first: $-(x^2 - 6x + 5) = -\big((x - 3)^2 - 4\big) = 4 - (x - 3)^2$. Getting this sign wrong flips which trig substitution you need.

Completing the square finds the vertex of any parabola, the resonant peak in a circuit's frequency response, and the mean and spread hidden inside a Gaussian $e^{-(ax^2 + bx)}$, which is why it shows up all over statistics.