Partial fractions in integration

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In integral calculus, partial fraction expansions provide an approach to integrating a general rational function. Any rational function of a real variable can be written as the sum of a polynomial function and a finite number of algebraic fractions. Each fraction in the expansion has as its denominator a polynomial function of degree 1 or 2, or some positive integer power of such a polynomial. (In the case of rational function of a complex variable, all denominators will have a polynomial of degree 1, or some positive integer power of such a polynomial.) If the denominator is a 1st-degree polynomial or a power of such a polynomial, then the numerator is a constant. If the denominator is a 2nd-degree polynomial or a power of such a polynomial, then the numerator is a 1st-degree polynomial.

Isaac Barrow's proof of the integral of the secant function was the earliest use of partial fractions in integration. In 1599, Edward Wright gave a solution by numerical methods – what today we would call Riemann sums.

A 1st-degree polynomial in the denominator

The substitution u = ax + b, du = a dx reduces the integral

$\int {1 \over ax+b}\,dx$

to

$\int {1 \over u}\,{du \over a}={1 \over a}\int{du\over u}={1 \over a}\ln\left|u\right|+C = {1 \over a} \ln\left|ax+b\right|+C.$

A repeated 1st-degree polynomial in the denominator

The same substitution reduces such integrals as

$\int {1 \over (ax+b)^8}\,dx$

to

$\int {1 \over u^8}\,{du \over a}={1 \over a}\int u^{-8}\,du = {1 \over a} \cdot{u^{-7} \over(-7)}+C = {-1 \over 7au^7}+C = {-1 \over 7a(ax+b)^7}+C.$

An irreducible 2nd-degree polynomial in the denominator

Next we consider such integrals as

$\int {x+6 \over x^2-8x+25}\,dx.$

The quickest way to see that the denominator x² − 8x + 25 is irreducible is to observe that its discriminant is negative. Alternatively, we can complete the square:

$x^2-8x+25=(x^2-8x+16)+9=(x-4)^2+9\,$

and observe that this sum of two squares can never be 0 while x is a real number.

In order to make use of the substitution

$\begin{align} u & = x^2-8x+25 \\ du & =(2x-8)\,dx \\ du/2 & = (x-4)\,dx \end{align}$

we would need to find x − 4 in the numerator. So we decompose the numerator x + 6 as (x − 4) + 10, and we write the integral as

$\int {x-4 \over x^2-8x+25}\,dx + \int {10 \over x^2-8x+25}\,dx.$

The substitution handles the first summand, thus:

$\int {x-4 \over x^2-8x+25}\,dx = \int {du/2 \over u} = {1 \over 2}\ln\left|u\right|+C = {1 \over 2}\ln(x^2-8x+25)+C.$

Note that the reason we can discard the absolute value sign is that, as we observed earlier, (x − 4)² + 9 can never be negative.

Next we must treat the integral

$\int {10 \over x^2-8x+25} \, dx.$

First, complete the square, then do a bit more algebra:

$\begin{align} & {} \quad \int {10 \over x^2-8x+25} \, dx = \int {10 \over (x-4)^2+9} \, dx \\[9pt] & = \int {10/9 \over \left({x-4 \over 3}\right)^2+1}\,dx = {10 \over 3} \int {1 \over \left({x-4 \over 3}\right)^2+1}\, \left({dx \over 3}\right) \end{align}$

Now the substitution

$w=(x-4)/3\,$

$dw=dx/3\,$

gives us

${10 \over 3}\int {dw \over w^2+1} = {10 \over 3} \arctan(w)+C={10 \over 3} \arctan\left({x-4 \over 3}\right)+C.$

Putting it all together,

$\int {x + 6 \over x^2-8x+25}\,dx = {1 \over 2}\ln(x^2-8x+25) + {10 \over 3} \arctan\left({x-4 \over 3}\right) + C.$

Using complex expanding

In some cases, having certain skill, it's more convenient to use the complex decomposition of the polynomial. So, in the example above:

$\int {x+6 \over x^2-8x+25}\,dx$

Expanding the denominator in the two complex multiplier:

x² − 8x + 25 = (x − 4 + 3i)(x − 4 − 3i)

Then looking for the expansion of the integrand into two terms

$\frac{x+6}{x^2-8x+25} = \frac{A}{x - 4 + 3i} + \frac{B}{x - 4 - 3i}$

Solving the system of linear equations, we obtain:

$A = \tfrac{1}{2} + \tfrac{5}{3}i, B = \tfrac{1}{2} - \tfrac{5}{3}i$

$\int \frac{x + 6}{x^2-8x+25}\,dx = ( \tfrac{1}{2} + \tfrac{5}{3}i) \int \frac{1}{x - 4 + 3i}\,dx + ( \tfrac{1}{2} - \tfrac{5}{3}i ) \int \frac{1}{x - 4 - 3i}\,dx$

After the obvious integration we have:

$\left(\tfrac{1}{2} + \tfrac{5}{3}i \right) \ln(x - 4 + 3i) + \left( \tfrac{1}{2} - \tfrac{5}{3}i \right) \ln (x - 4 - 3i) + C$

Grouping the separate real and imaginary terms:

$\tfrac{1}{2} \left( \ln(x - 4 + 3i) + \ln(x - 4 - 3i) \right) + \tfrac{5}{3}i \left( \ln(x - 4 + 3i) - \ln(x - 4 - 3i) \right) + C$

$\tfrac{1}{2} \ln \left( (x - 4 + 3i)(x - 4 - 3i) \right) + \tfrac{5}{3}i \ln \frac{x - 4 + 3i}{x - 4 - 3i} + C$

$\tfrac{1}{2} \ln (x^2-8x+25) + \tfrac{5}{3}i \ln \frac{1 - i \frac{x - 4}{3}}{1 + i \frac{x - 4}{3}} + C$

As it's known, the arctangent of a complex variable can be expressed by the logarithm:

$\arctan \, z = \tfrac{1}{2}i \ln \frac{1-i\,z}{1+i\,z}$

This allows us to express the second term in the arctangent:

$\tfrac{1}{2} \ln (x^2-8x+25) + \tfrac{10}{3} \arctan \frac{x - 4}{3} + C$

A repeated irreducible 2nd-degree polynomial in the denominator

Next, consider

$\int {x+6 \over (x^2-8x+25)^{8}}\,dx.$

Just as above, we can split x + 6 into (x − 4) + 10, and treat the part containing x − 4 via the substitution

$\begin{align} u & = x^2-8x+25, \\ du & = (2x-8)\,dx, \\ du/2 & = (x-4)\,dx. \end{align}$

This leaves us with

$\int {10 \over (x^2-8x+25)^{8}}\,dx.$

As before, we first complete the square and then do a bit of algebraic massaging, to get

$\int {10 \over (x^2-8x+25)^{8}}\,dx =\int {10 \over ((x-4)^2+9)^{8}}\,dx =\int {10/9^{8} \over \left(\left({x-4 \over 3}\right)^2+1\right)^8}\,dx.$

Then we can use a trigonometric substitution:

$\tan\theta={x-4 \over 3},\,$

$\left({x-4 \over 3}\right)^2+1=\tan^2\theta+1=\sec^2\theta,\,$

$d\tan\theta =\sec^2\theta\,d\theta={dx \over 3}.\,$

Then the integral becomes

$\int {30/9^{8} \over \sec^{16}\theta} \sec^2\theta \,d\theta ={30 \over 9^{8}}\int \cos^{14} \theta \, d\theta.$

By repeated applications of the half-angle formula

$\cos^2\theta={1 \over 2}( 1 + \cos(2\theta)),$

one can reduce this to an integral involving no higher powers of cos θ higher than the 1st power.

Then one faces the problem of expression sin(θ) and cos(θ) as functions of x. Recall that

$\tan(\theta)={x - 4 \over 3},$

and that tangent = opposite/adjacent. If the "opposite" side has length x − 4 and the "adjacent" side has length 3, then the Pythagorean theorem tells us that the hypotenuse has length √((x − 4)² + 3²) = √(x² −8x + 25).

Therefore we have

$\sin(\theta) = {\text{opposite} \over \text{hypotenuse}} = {x-4 \over \sqrt{x^2 - 8x + 25}},$

$\cos(\theta) = {\text{adjacent} \over \text{hypotenuse}} = {3 \over \sqrt{x^2 - 8x + 25}},$

and

$\sin(2\theta) = 2\sin(\theta)\cos(\theta) = {6(x-4) \over x^2 - 8x + 25}.$

External links

• Partial Fraction Expander
• Mathematical Assistant on Web online calculation of integrals, allows to integrate in small steps (includes partial fractions, powered by Maxima (software))