# Computing Integrals by Completing the Square

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Computing Integrals by Completing the Square – HMC Calculus Tutorial

We will review the method of completing the square in the context of evaluating integrals:

###### Example

Let’s start by evaluating $\int\frac{dx}{2x^2-12x+26}.$ The denominator does not factor with rational coefficients, so partial fractions is not a viable option. There is also no obvious substitution to make. Instead, we will complete the square in the denominator to get a recognizable form for the integral.

Now \begin{eqnarray*} 2x^2-12x+26&=&2[x^2-6x+13]\\ &=&2[(x^2-6x+9)+4]\\ &=&2[(x-3)^2+4]. \end{eqnarray*} Returning to the integral, \begin{eqnarray*} \int\frac{dx}{2x^2-12x+26}&=&\int\frac{dx}{2[(x-3)^2+4]}\\ &=&\frac{1}{2}\int\frac{dx}{(x-3)^2+2^2}\\ &=&\frac{1}{2}\left[\frac{1}{2}\arctan\left(\frac{x-3}{2}\right)\right]+C\\ &=&\frac{1}{4}\arctan\left(\frac{x-3}{2}\right)+C. \end{eqnarray*}

Certain other types of integrals can be evaluated by this method as well:

###### Example

Consider $\int \frac{dx}{\sqrt{21-4x-x^2}}.$ Now \begin{eqnarray*} 21-4x-x^2&=&21-[x^2+4x]\\ &=&21+4-[x^2+4x+4]\\ &=&25-(x+2)^2. \end{eqnarray*} Returning to the integral, \begin{eqnarray*} \int \frac{dx}{\sqrt{21-4x-x^2}}&=&\int \frac{dx}{\sqrt{25-(x+2)^2}}\\ &=& \arcsin \left(\frac{x+2}{5}\right)+C. \end{eqnarray*}

Completing the square is a powerful method that is used to derive the quadratic formula:

We will find the roots of $ax^2+bx+c=0$: \begin{eqnarray*} ax^2+bx+c&=&0\\ x^2+\frac{b}{a}x+\frac{c}{a}&=&0\\ x^2+\frac{b}{a}x\qquad&=&-\frac{c}{a}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}&=&\frac{b^2}{4a^2}-\frac{c}{a}\\ \left(x+\frac{b}{2a}\right)^2&=&\frac{b^2-4ac}{4a^2}\\ x+\frac{b}{2a}&=&\pm \frac{\sqrt{b^2-4ac}}{2a}\\ x&=&\frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{eqnarray*} which is the familiar quadratic formula!

#### Key Concept

By completing the square, we may rewrite any quadratic polynomial $ax^2+bx+x$ in the form $a\left[(x+k_1)^2+k_2\right]$ where $k_1$ and $k_2$ may be positive or negative. Integrals containing negative or non-integer powers of $ax^2+bx+c$ can often be computed using a trigonometric substitution or looked up in an integral table after being rewritten in this form.