Integration Using Completing the Square (College Board AP® Calculus AB): Revision Note

Integration using completing the square

What is completing the square?

• Completing the square is a way of rewriting quadratic expressions

  • A quadratic expression is of the form $a{x}^2+bx+c$

  • Rewriting can make them easier to use for various purposes

• The simple version of completing the square is

  • $x^2+bx+c={\left(x+\frac{b}{2}\right)}^2+c-{\left(\frac{b}{2}\right)}^2$

    • E.g. $x^2-6x+2={\left(x-3\right)}^2+2-{\left(-3\right)}^2={\left(x-3\right)}^2-7$

• When there is a coefficient in front of the ${\mathit{x}}^{\mathbf{2}}$ this becomes

  • $a{x}^2+bx+c=a{\left(x+\frac{b}{2a}\right)}^2+c-a{\left(\frac{b}{2a}\right)}^2$

    • E.g. $3x^2+12x-7=3{\left(x+2\right)}^2-7-3{\left(2\right)}^2=3{\left(x+2\right)}^2-19$

How can I integrate using completing the square?

• Recall the standard integrals

  • $\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin x+C,-1<x<1$

  • $\int \frac{1}{1+x^2}dx=\mathrm{arc}\tan x+C$

  • See the 'Derivatives & Antiderivatives' study guide

• Completing the square can be used to get integrals into a form where these results can be used

  • The methods from the 'Integrals of Composite Functions' study guide may be needed as well

• E.g. $\int \frac{1}{x^2-6x+13}dx$

  • Complete the square on the denominator

    • $x^2-6x+13={\left(x-3\right)}^2+4$

  • Rewrite the function to be integrated

    • $\frac{1}{x^2-6x+13}=\frac{1}{4+{\left(x-3\right)}^2}=\frac{1}{4}·\frac{1}{1+{\left({\displaystyle \frac{x-3}{2}}\right)}^2}$

  • That is now a multiple of $\frac{1}{1+{(\dots )}^2}$

    • This tells us that $\arctan (\dots )$ is going to be be our likely answer

  • To see if 'adjust and compensate' is needed, use the chain rule to differentiate $\arctan \left(\frac{x-3}{2}\right)$

    • $\frac{d}{dx}\left(\arctan \left(\frac{x-3}{2}\right)\right)=\frac{1}{1+{\left({\displaystyle \frac{x-3}{2}}\right)}^2}·\frac{1}{2}$

    • $\frac{d}{dx}\left(\frac{1}{2}\arctan \left(\frac{x-3}{2}\right)\right)=\frac{1}{1+{\left({\displaystyle \frac{x-3}{2}}\right)}^2}·\frac{1}{4}$

  • Integrate

    • $\int \frac{1}{x^2-6x+13}dx=\frac{1}{2}\arctan \left(\frac{x-3}{2}\right)+C$