# Evaluate Integrals Involving Quadratic Expressions Using Completing Square

Tutorials with examples and detailed solutions and exercises with answers on how to use the techniques of completing square and substitution to evaluate integrals involving quadratic expressions.

## Formulas

We first review some of the derivatives formulas for known inverse functions involving quadratic expressions.
We now use the above differentiation formulas to write integrals as follows.

NOTE: in what follows, K is the constant of integration. Finally one more integral is computed using partial fractions decomposition as follows
5)
\int \dfrac{1}{{1 - x^2}} dx = \int (\dfrac{1/2}{x+1}- \dfrac{1/2}{x-1}) dx = \dfrac{1}{2} \ln \dfrac{x+1}{x-1} + K

## Examples

### 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 - x2 - x as follows
- x2 - x = - ( x2 + x ) = - (x + 1/2)2 + 1/4
The given integral may written as follows

= \int \dfrac{1}{\sqrt{1/4 - (x+1/2)^2}} dx =

Factor out 1/4 from under the square root
= \int \dfrac{2}{\sqrt{1 - (2(x+1/2))^2}} dx

Let z = 2(x + 1/2) = 2x + 1 and therefore dz/2 = dx and the integral becomes
= \int \dfrac{2}{\sqrt{1 - z^2}} dz

= arcsin(z) + K
= arcsin(2x + 1) + K

### 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 3x2 + 12x + 24 as follows
3x2 + 12x + 24 = 3( x2 + 4x) + 24
= 3( (x + 2)2 - 22) + 24
= 3(x + 2)2 + 12
Factor 12 out and rewrite as follows
= 12 [ ((1/2)(x + 2))2 + 1]
The given integral may written as follows

= (1 / 6) \int \dfrac{1}{(x/2 + 1)^2 + 1} dx

Let z = x/2 + 1 and therefore dx = 2dz and rewrite the integral as
= (1 / 3) \int \dfrac{1}{z^2 + 1} dz
= (1/3) arctan(z) + K
= (1/3) arctan(x/2 + 1) + K

### Example 3

Evaluate the integral
\int \dfrac{1}{\sqrt{x^2 + 12x + 40}} dx

Solution to Example 3:
We first complete the square for the expression x2 + 12x + 40 as follows
x2 + 12x + 40 = ( x + 6 )2 + 4
The given integral may written as follows

= \int \dfrac{1}{\sqrt{( x + 6 )^2 + 4}} dx

Factor 4 out from under the square root
= \int \dfrac{1}{2\sqrt{( x/2+ 3 )^2 + 1}} dx
Let z = x/2 + 3, hence 2 dz = dx, and the integral may be written
= \int \dfrac{1}{\sqrt{z^2 + 1}} dz

= arcsinh(z) + K
= arcsinh(x/2 + 3) + K

## Exercises

Evaluate the integrals given below
1.
\int \dfrac{3}{\sqrt{9 - x^2}} dx

2.
\int \dfrac{3}{x^2 + 12x + 45} dx

3.
\int \dfrac{\sqrt2}{\sqrt{2x^2 + 10x + 13}} dx