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How to Integrate 1/(ax^2 + bx + c) -- Antiderivative of a Quadratic Rational Function

Updated on February 21, 2014
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TR Smith is a product designer and former teacher who uses math in her work every day.

The function f(x) = 1/(ax^2 + bx + c) is a rational function with a quadratic denominator. There are three cases to consider for the quadratic ax^2 + bx + c with real number coefficients: the expression has two distinct real roots, the equation has one real root repeated, or the equation has two complex roots. Depending on the case, the method of integration is different.

If the quadratic expression has two distinct real roots, then you find the antiderivative of 1/(ax^2 + bx + c) by factoring the quadratic and using the method of partial fractions. If the expression has one repeated root, then the expression is a perfect square and can be integrated the same as 1/x^2. And finally, if the expression has no real roots, then you complete the square to put it in the form x^2 + r^2. Each of these cases is worked out with an example.


Example 1: Antiderivative of 1/(2x^2 + 12x + 10)

The quadratic 2x^2 + 12x + 10 factors as 2(x+1)(x+5), which has the roots x = -1 and x = -5. Since these are distinct real numbers, we must simplifly the integrand 1/[2(x+1)(x+5)] using the technique of partial fractions. The first thing we will do is factor out the 1/2 and just look at the rational function 1/[(x+1)(x+5)]. What we want to do is find some constants A and B such that

A/(x+1) + B/(x+5) = 1/[(x+1)(x+5)]

Giving both sides a like denominator produces the equivalent equation

A(x+5)/[(x+1)(x+5)] + B(x+1)/[(x+1)(x+5)] = 1/[(x+1)(x+5)]

Clearing the denominator gives us

A(x+5) + B(x+1) = 1
(A+B)x + (5A+B) = 0x + 1

This gives us two equations in two unknown variables A and B:

A + B = 0 and 5A + B = 1

Using methods of solving linear systems, we see that A = 1/4 and B = -1/4. Putting the factor of 1/2 back in gives us the equivalence

1/[2(x+1)(x+5)] = (1/8)/(x+1) + (-1/8)/(x+5)

And now we can compute the integral of 1/[2(x+1)(x+5)] as follows:

∫ 1/[2(x+1)(x+5)] dx
= ∫ (1/8)/(x+1) dx + ∫ (-1/8)/(x+5) dx
= (1/8)Ln(x+1) - (1/8)Ln(x+5) + C
= (1/8)Ln[ (x+1)/(x+5) ] + C

Example 2: Antiderivative of 1/(2x^2 + 12x + 18)

The quadratic expression 2x^2 + 12x + 18 factors into 2(x+3)(x+3) = 2(x+3)^2, which has the repeated real root x = -3. This case can be integrated using a simple u-substitution. First we have

∫ 1/[2x^2 + 12x + 18] dx
= (1/2) * ∫ 1/(x+3)^2 dx

Now we use the substitution x = u - 3 with dx = du. This gives us

(1/2) * ∫ 1/(x+3)^2 dx
= (1/2) * ∫ 1/u^2 du
= (1/2) * -1/u + C
= -1/(2u) + C

And now the reverse substitution gives us

-1/(2u) + C
= -1/(2x + 6) + C

Example 3: Antiderivative of 1/(2x^2 + 12x + 26)

Out last example has the quadratic expression 2x^2 + 12x + 26 in the denominator, but this polynomial has no real roots. So instead of factoring it, we complete the square.

2x^2 + 12x + 26
= 2[x^2 + 6x + 13]
= 2[x^2 + 6x + 9 + 4]
= 2[(x^2 + 6x + 9) + 4]
= 2[(x+3)^2 + 2^2]

This gives us the integral

∫ 1/[2x^2 + 12x + 26] dx
= (1/2) * ∫ 1/[(x+3)^2 + 2^2] dx

Now we use the substitution x + 3 = 2u and dx = 2 du. This gives us

(1/2) * ∫ 1/[(2u)^2 + 2^2] * 2 du
= ∫ 1/[(2u)^2 + 2^2] du
= (1/4) * ∫ 1/[u^2 + 1] du
= (1/4)*arctan(u) + C

The reverse substitution gives us

(1/4)*arctan(u) + C
= (1/4)*arctan[(x+3)/2] + C

Comparison of the Three Integrals

The three functions f(x) = 1/[2x^2 + 12x + 10], g(x) = 1/[2x^2 + 12x + 18], and h(x) = 1/[2x^2 + 12x + 26] are almost identical except for the constant terms in the denominators -- 10, 18, and 26. Yet such a seemingly trivial aspect of the function creates very different integrals. The antiderivative of f(x) is a logarithmic function, while that of g(x) is a simple rational function, and that of h(x) is an inverse trigonometric function.

Below are the graphs of f(x), g(x), h(x), and their antiderivatives. The graph of f(x) has two vertical asymptotes, while g(x) has one, and h(x) has none.

The graphs of f(x) = 1/(2x^2+12x+10) in red, g(x) = 1/(2x^2+12x+18) in blue, and h(x) = 1/(2x^2+12x+26) in green. Notice x = -3 is the line of symmetry for all three.
The graphs of f(x) = 1/(2x^2+12x+10) in red, g(x) = 1/(2x^2+12x+18) in blue, and h(x) = 1/(2x^2+12x+26) in green. Notice x = -3 is the line of symmetry for all three.
Graphs of the antiderivatives of f(x) in red, g(x) in blue, and h(x) in green.
Graphs of the antiderivatives of f(x) in red, g(x) in blue, and h(x) in green.

What if the numerator isn't 1?

To integrate a function such as (x^3 + 5x^2 - 17x + 11)/(x^2 + 8x + 4), where the numerator is a polynomial, you break up the numerator so that you can express the original rational function as a sum of simpler rational functions that are easier to integrate. Breaking down the numerator gives us

(x^3 + 5x^2 - 17x + 11)

= (x^3 + 8x^2 + 4x) + (-3x^2 - 24x - 12) + (3x + 12) + 11

Putting this over the denominator x^2 + 8x + 4 gives us four integrals that are easier to work with.

∫ (x^3 + 8x^2 + 4x)/(x^2 + 8x + 4) dx
= ∫ x dx
= (1/2)x^2 + C

∫ (-3x^2 - 24x - 12)/(x^2 + 8x + 4) dx
= ∫ -3 dx
= -3x + C

∫ (3x + 12)/(x^2 + 8x + 4) dx
u = (x^2 + 8x + 4) and du = 2x + 8 dx
= ∫ (3/2) du / u
= (3/2)Ln(u)
= (3/2)Ln(x^2 + 8x + 4) + C

∫ 11/(x^2 + 8x + 4) dx
= ∫ 11/[(x + p)(x + q)] dx

For the last integral, since x^2 + 8x + 4 has two real roots, you proceed using partial fractions. The complete antiderivative of (x^3 + 5x^2 - 17x + 11)/(x^2 + 8x + 4) is shown in the image below.

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