E4 - Partial fractions with quadratic factors

E4 - Partial fractions with quadratic factors

Partial fractions can split a rational integrand even when a quadratic factor has no real roots. You will choose the right numerator over each factor, find its coefficients, then separate logarithmic and inverse tangent integrals.

Irreducible quadratic factors

How can you integrate 2x2+3(x−1)(x2+4)\displaystyle\frac{2x^2+3}{(x-1)(x^2+4)} when x2+4x^2+4 will not factorise over the real numbers? It is a proper rational function, so partial fractions can still simplify it. The new feature is the numerator placed over the quadratic.

Compare x2−4=(x−2)(x+2)x^2-4=(x-2)(x+2) with x2+4x^2+4. The first has two real linear factors. The second is positive for every real xx, so it has no real linear factors.

Irreducible quadratic

A quadratic is irreducible over the real numbers if it cannot be written as a product of real linear factors. For ax2+cax^2+c, with a≠0a\ne0, this occurs when ac>0ac>0. If ac<0ac<0, it has two real roots; if c=0c=0, it contains a repeated factor xx.

For integration we can take a>0a>0, c>0c>0, pulling out an overall minus sign first if both are negative. Thus 4x2+94x^2+9 is irreducible, whereas 4x2−9=(2x−3)(2x+3)4x^2-9=(2x-3)(2x+3) is not. A constant expression with a=0a=0 is not a quadratic.

Match the numerator to the factor

A proper rational function has numerator degree smaller than denominator degree. If it is improper, first use polynomial division. Then use a numerator of degree lower than each denominator factor:

Factor in the denominatorCorresponding partial-fraction term
A single linear factor x−rx-rA/(x−r)A/(x-r)
An irreducible quadratic ax2+cax^2+c(Bx+C)/(ax2+c)(Bx+C)/(ax^2+c)
A repeated linear factor (x−r)2(x-r)^2A/(x−r)+B/(x−r)2A/(x-r)+B/(x-r)^2

The quadratic numerator needs both a linear term and a constant. These are unknown real coefficients; a solution may make either coefficient zero. Assuming a constant numerator in advance can make the coefficient equations impossible.

A repeated irreducible quadratic follows the same all-powers rule. If (ax2+c)2(ax^2+c)^2 occurs, include

Ax+Bax2+c+Cx+D(ax2+c)2\frac{Ax+B}{ax^2+c}+\frac{Cx+D}{(ax^2+c)^2}

This states the decomposition pattern; the two terms need not both be nonzero. Do not omit the lower power. The same rule includes every power up to the multiplicity for a more highly repeated factor.

For example, the complete form for a repeated linear factor together with a quadratic is

1(x−1)2(x2+4)=Ax−1+B(x−1)2+Cx+Dx2+4\frac{1}{(x-1)^2(x^2+4)} =\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+4}

This is an identity for allowed xx, not a claim that the unknown coefficients equal 1. Multiplying by the common denominator is what will determine them.

Key idea

An irreducible quadratic stays together. Give it a general linear numerator; for a repeated factor include every denominator power, each with the appropriate numerator.

Choose the complete form before doing any coefficient calculation.

Finding the linear numerator

To find the coefficients, clear the fractions and compare coefficients of matching powers of xx. A real root of a linear factor can isolate one constant. An irreducible quadratic has no real root to substitute, so coefficient comparison supplies the remaining equations.

For the opening fraction, start with

2x2+3(x−1)(x2+4)=Ax−1+Bx+Cx2+4\frac{2x^2+3}{(x-1)(x^2+4)} =\frac{A}{x-1}+\frac{Bx+C}{x^2+4}

This form contains three unknowns, enough to match the three coefficients of a quadratic identity. The decomposition is algebra, so it is not supplied as a formula in the booklet.

Worked example 1: a linear factor and an irreducible quadratic

Question. Express 2x2+3(x−1)(x2+4)\displaystyle\frac{2x^2+3}{(x-1)(x^2+4)} in partial fractions.

Plan. Use a constant over the linear factor and a linear numerator over the quadratic. Clear denominators, isolate AA, then compare coefficients.

2x2+3=A(x2+4)+(Bx+C)(x−1)polynomial identityx=1:5=5Aisolate AA=12x2+3=(1+B)x2+(C−B)x+(4−C)expand the identityB=1,C=1compare x2 and x\begin{aligned} 2x^2+3&=A(x^2+4)+(Bx+C)(x-1) && \text{polynomial identity}\\ x=1:\quad5&=5A && \text{isolate }A\\ A&=1\\ 2x^2+3&=(1+B)x^2+(C-B)x+(4-C) && \text{expand the identity}\\ B&=1,\qquad C=1 && \text{compare }x^2\text{ and }x \end{aligned}

The constant equation 4−C=34-C=3 agrees. We may put x=1x=1 into the polynomial identity even though the original fraction is undefined there: after clearing denominators, the identity holds as a polynomial.

Answer. 1x−1+x+1x2+4\displaystyle\frac1{x-1}+\frac{x+1}{x^2+4}, for x≠1x\ne1.

Check. Recombine the proposed terms. Their numerator is x2+4+(x+1)(x−1)=x2+4+x2−1=2x2+3x^2+4+(x+1)(x-1)=x^2+4+x^2-1=2x^2+3.

Change only the linear denominator factor from x−1x-1 to x+1x+1. The constant term of the quadratic numerator will change sign.

Worked example 2: changing the linear factor

Question. Express 2x2+3(x+1)(x2+4)\displaystyle\frac{2x^2+3}{(x+1)(x^2+4)} in partial fractions.

Plan. Keep the same numerator form, but use x=−1x=-1 to isolate the coefficient over x+1x+1.

2x2+3=A(x2+4)+(Bx+C)(x+1)clear denominatorsx=−1:5=5A⇒A=12x2+3=(1+B)x2+(B+C)x+(4+C)expandB=1,C=−1compare coefficients\begin{aligned} 2x^2+3&=A(x^2+4)+(Bx+C)(x+1) && \text{clear denominators}\\ x=-1:\quad5&=5A\quad\Rightarrow\quad A=1\\ 2x^2+3&=(1+B)x^2+(B+C)x+(4+C) && \text{expand}\\ B&=1,\qquad C=-1 && \text{compare coefficients} \end{aligned}

Answer. 1x+1+x−1x2+4\displaystyle\frac1{x+1}+\frac{x-1}{x^2+4}, for x≠−1x\ne-1.

Check. Recombining gives numerator x2+4+(x−1)(x+1)=2x2+3x^2+4+(x-1)(x+1)=2x^2+3, exactly as required.

For repeated factors, the same process works after writing the complete form from the previous part. Clearing denominators produces a polynomial identity, and you compare enough coefficients to find all the unknowns.

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