Evaluate Integrals by Completing the Square

Review Formulas, Step-by-Step Worked Examples with Solutions Closed, Exercises, and References

Examples, detailed solutions, and exercises with answers on how to use the techniques of completing the square and substitution to evaluate integrals involving quadratic expressions. Exercises with their answers are also included.

Review Formulas of Derivatives and Integrals

NOTE: In what follows, \( c \) is the constant of integration.

We first review some of the derivative formulas for known inverse functions involving quadratic expressions:

\[ \begin{aligned} & \dfrac{d}{dx} \arcsin x = \dfrac{1}{\sqrt{1 - x^2}} \\[15pt] & \dfrac{d}{dx} \arctan x = \dfrac{1}{1 + x^2} \\[15pt] & \dfrac{d}{dx} \operatorname{arcsinh} x = \dfrac{1}{\sqrt{1 + x^2}} \\[15pt] & \dfrac{d}{dx} \operatorname{arccosh} x = \dfrac{1}{\sqrt{x^2 - 1}} \end{aligned} \]

We now use the above differentiation formulas to write the corresponding integral formulas:

\[ \begin{aligned} & \int \dfrac{1}{\sqrt{1 - x^2}} \, dx = \arcsin x + c \\[15pt] & \int \dfrac{1}{1 + x^2} \, dx = \arctan x + c \\[15pt] & \int \dfrac{1}{\sqrt{1 + x^2}} \, dx = \operatorname{arcsinh} x + c \\[15pt] & \int \dfrac{1}{\sqrt{x^2 - 1}} \, dx = \operatorname{arccosh} x + c \end{aligned} \]

Examples with Detailed Solutions

Example 1

Evaluate the integral:

\[ \int \dfrac{1}{\sqrt{-x^2 - x}} \, dx \]
Solution to Example 1

We first complete the square for the expression \( -x^2 - x \):

\[ \begin{aligned} & -x^2 - x \\ & = -(x^2 + x) \\ & = -\left( \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} \right) \\ & = \dfrac{1}{4} - \left(x + \frac{1}{2}\right)^2 \end{aligned} \]

Substitute the above into the given integral:

\[ \int \dfrac{1}{\sqrt{-x^2 - x}} \, dx = \int \dfrac{1}{\sqrt{\dfrac{1}{4} - \left(x + \frac{1}{2}\right)^2}} \, dx \]

Factor out \( \dfrac{1}{4} \) from under the square root:

\[ = \int \dfrac{1}{\sqrt{\dfrac{1}{4}\left(1 - \left(2\left(x + \frac{1}{2}\right)\right)^2\right)}} \, dx = \int \dfrac{2}{\sqrt{1 - (2x + 1)^2}} \, dx \]

Use the method of substitution: let \( z = 2x + 1 \), so \( dz = 2 \, dx \) (or \( dx = \dfrac{dz}{2} \)):

\[ = \int \dfrac{1}{\sqrt{1 - z^2}} \, dz = \arcsin(z) + c \]

Substitute back \( z = 2x + 1 \) to obtain the final answer:

\[ \int \dfrac{1}{\sqrt{-x^2 - x}} \, dx = \arcsin(2x + 1) + c \]

Example 2

Evaluate the integral:

\[ \int \dfrac{2}{3x^2 + 12x + 24} \, dx \]
Solution to Example 2

We first complete the square for the expression \( 3x^2 + 12x + 24 \):

\[ \begin{aligned} & = 3(x^2 + 4x) + 24 \\ & = 3\left((x + 2)^2 - 4\right) + 24 \\ & = 3(x + 2)^2 - 12 + 24 = 3(x + 2)^2 + 12 \\ & = 12\left(\dfrac{1}{4}(x + 2)^2 + 1\right) = 12\left(\left(\dfrac{x}{2} + 1\right)^2 + 1\right) \end{aligned} \]

Substitute into the integral and use substitution (\( z = \dfrac{x}{2} + 1 \), \( dx = 2 \, dz \)):

\[ \int \dfrac{2}{3x^2 + 12x + 24} \, dx = \int \dfrac{2}{12\left(\left(\dfrac{x}{2} + 1\right)^2 + 1\right)} \, dx = \dfrac{1}{6} \int \dfrac{1}{z^2 + 1} (2 \, dz) = \dfrac{1}{3} \int \dfrac{1}{z^2 + 1} \, dz \]

Evaluate and substitute back:

\[ = \dfrac{1}{3} \arctan(z) + c = \dfrac{1}{3} \arctan\left(\dfrac{x}{2} + 1\right) + c \]

Example 3

Evaluate the integral:

\[ \int \dfrac{1}{\sqrt{x^2 + 12x + 40}} \, dx \]
Solution to Example 3

Complete the square for \( x^2 + 12x + 40 \):

\[ x^2 + 12x + 40 = (x + 6)^2 - 36 + 40 = (x + 6)^2 + 4 \]

Rewrite the integral:

\[ \int \dfrac{1}{\sqrt{(x + 6)^2 + 4}} \, dx = \int \dfrac{1}{\sqrt{4\left(\left(\dfrac{x}{2} + 3\right)^2 + 1\right)}} \, dx = \int \dfrac{1}{2\sqrt{\left(\dfrac{x}{2} + 3\right)^2 + 1}} \, dx \]

Let \( z = \dfrac{x}{2} + 3 \), so \( dx = 2 \, dz \):

\[ = \int \dfrac{1}{\sqrt{z^2 + 1}} \, dz = \operatorname{arcsinh}(z) + c = \operatorname{arcsinh}\left(\dfrac{x}{2} + 3\right) + c \]

Example 4

Evaluate the integral:

\[ \int \dfrac{1}{10 + x^2 - 2x} \, dx \]
Solution to Example 4

Complete the square for the denominator \( 10 + x^2 - 2x \):

\[ x^2 - 2x + 10 = (x - 1)^2 - 1 + 10 = (x - 1)^2 + 9 \]

Rewrite the integral and factor out \( 9 \):

\[ \int \dfrac{1}{(x - 1)^2 + 9} \, dx = \int \dfrac{1}{9\left(\left(\dfrac{x - 1}{3}\right)^2 + 1\right)} \, dx \]

Let \( z = \dfrac{x - 1}{3} \), so \( dx = 3 \, dz \):

\[ = \dfrac{1}{9} \int \dfrac{1}{z^2 + 1} (3 \, dz) = \dfrac{1}{3} \int \dfrac{1}{z^2 + 1} \, dz = \dfrac{1}{3} \arctan(z) + c \]

Substitute back \( z \):

\[ = \dfrac{1}{3} \arctan\left(\dfrac{x - 1}{3}\right) + c \]

Exercises

Evaluate the integrals given below:

  1. \( \displaystyle \int \dfrac{3}{\sqrt{9 - x^2}} \, dx \)
  2. \( \displaystyle \int \dfrac{3}{x^2 + 12x + 45} \, dx \)
  3. \( \displaystyle \int \dfrac{\sqrt{2}}{\sqrt{2x^2 + 10x + 13}} \, dx \)
  4. \( \displaystyle \int \dfrac{1}{5 + x^2 + 2x} \, dx \)

Answers to Above Exercises

  1. \( 3 \arcsin\left(\dfrac{x}{3}\right) + c \)
  2. \( \arctan\left(\dfrac{x}{3} + 2\right) + c \)
  3. \( \operatorname{arcsinh}(2x + 5) + c \)
  4. \( \dfrac{1}{2} \arctan\left(\dfrac{x + 1}{2}\right) + c \)

More References and Links

  1. Integrals and their applications in calculus
  2. Completing the square tutorial
  3. Integration by substitution tutorial