Integration of Quadratic Expressions

Intro

An irreducible quadratic has no factors.
When the integrand includes an irreducible quadratic in the denominator, we
complete the square and use substitution. Since completing the square turns
the irreducible quadratic into a sum or difference of squares, we sometimes use
trig substitutions to solve the integral especially when the quadratic is square rooted.

.

Example 1

Evaluate

When we complete the square on the quadratic denominator we get
x² – 6x + 9 – 9 + 13 = (x 3)² + 4, so now the integral is

so we use a substitution

Let u = x – 3, so du = dx and x = u + 3

This last statement means that 2x – 1 = 2(u + 3) – 1 = 2u + 5

Now our integral is

Using the substitution w = u² + 4, for the 1st integral, we get

ln (u² + 4) + (5/2) arcTan (u/2) and now we "unsubstitute"

The final answer is

_____________________

Example 2

Evaluate

To complete the square here, we first factor out a negative from the x terms, since
every perfect square includes a positive x².

So 8 + 2x x 2 = 8 – (x² – 2x) = 8 – (x² – 2x + 1) + 1 = 9 – (x – 1)².

Using the substitution u = x – 1 and du = dx, the integral becomes

and this integrates to .

______________________________________

Example 3

Now we do one which requires a trig substitution to solve.

Evaluate

When we complete the square on the trinomial we get (x + 4)² – 16 + 25 = (x + 4)² + 9.

Using the substitution u = x + 4 and du = dx, the integral becomes

Now, with the trig sub u = 3 tan A and du = 3 sec² A dA , we get

and since 1 + tan² A = sec² A

we get

To get back to x's, the original variable we'll use the triangle created by the
substitution u = 3 tan A or u/3 = tan A. We'll also replace u with x + 4.

The answer then is

Practice

Evaluate these integrals by completing the square on the irreducible quadratic.

 1) 2) 3) 4) 5) Find the area bounded by the graphs of , y = 0, x = 0, and x = 1. hint: use the same technique as for #4. 6) The velocity at time t of a point moving along a coordinate line is How far does the point travel during the interval t = 0 to t = 5 seconds?

Solutions

 1) x² – 4x + 8 = (x – 2)² + 4.so the integral I = I = ½ arcTan [½(x – 2)] + C 2) 7 + 6x – x² = 16 – (x – 3)²so the integral I =I = arcSin [¼(x – 3)] + C 3) 9 – 8x – x² = 25 – (x + 4)²so I =When we break this up into 2 integrals,I =I = 4) x³ – 1 = (x – 1)(x² + x + 1), we complete the square on the quadratic x² + x + 1. x² + x + 1 = (x + ½)² + ¾, now we set u = x + ½ and du = dx.This makes (x – 1) = (u – 3/2), so I =.Now we use partial fraction decomposition to solve the integrals.I = 5) Using the technique of #4, we get Note we changed the limits of integration to values of u.square units. .6)

(all content © MathRoom Learning Service; 2004 - ).