Partial fractions in integration

In integral calculus, the use of partial fractions is required to integrate the 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 partial fractions. Each partial fraction has as its denominator a polynomial function of degree 1 or 2, or some positive integer power of such a function. 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.

For an account of how to find this partial fraction expansion of a rational function, see partial fraction.

This article is about what to do "after" finding the partial fraction expansion, when one is trying to find the function's antiderivative.

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

:$int\; \{1\; over\; u\},\{du\; over\; a\}=\{1\; over\; a\}int\{duover\; u\}=\{1\; over\; a\}lnleft|u\; ight|+C\; =\; \{1\; over\; a\}\; lnleft|ax+b\; ight|+C.$

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 "x"² − 8"x" + 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

:$u=x^2-8x+25,$:$du=(2x-8),dx$:$du/2=(x-4),dx$

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:

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

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:

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

Now the substitution

:$w=(x-4)/3,$:$dw=dx/3,$

gives us

:$\{10\; over\; 3\}int\; \{dw\; over\; w^2+1\}=\; \{10\; over\; 3\}\; arctan(w)+C=\{10\; over\; 3\}\; arctanleft(\{x-4\; over\; 3\}\; ight)+C.$

Putting it all together,

:$int\; \{x\; +\; 6\; over\; x^2-8x+25\},dx=\; \{1\; over\; 2\}ln(x^2-8x+25)\; +\; \{10\; over\; 3\}\; arctanleft(\{x-4\; over\; 3\}\; ight)\; +\; C.$

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

:$u=x^2-8x+25,,$:$du=(2x-8),dx$:$du/2=(x-4),dx.$

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

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

Then we can use a trigonometric substitution:

:$an\; heta=\{x-4\; over\; 3\},,$

:$left(\{x-4\; over\; 3\}\; ight)^2+1=\; an^2\; heta+1=sec^2\; heta,,$

:$an\; heta\; ,d\; heta=sec^2\; heta,d\; heta=\{dx\; over\; 3\}.,$

Then the integral becomes

:$int\; \{30/9^\{8\}\; over\; sec^\{16\}\; heta\}\; sec^2\; heta\; ,d\; heta=\{30\; over\; 9^\{8int\; cos^\{14\}\; heta\; ,\; d\; heta$

By repeated applications of the half-angle formula

:$cos^2\; heta=\{1\; over\; 2\}(\; 1\; +\; cos(2\; heta)),$

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

:$an(\; heta)=\{x\; -\; 4\; over\; 3\},$

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) = √("x"² −8"x" + 25).

Therefore we have

:$sin(\; heta)\; =\; \{mathrm\{opposite\}\; over\; mathrm\{hypotenuse\; =\; \{x-4\; over\; sqrt\{x^2\; -\; 8x\; +\; 25,$

:$cos(\; heta)\; =\; \{mathrm\{adjacent\}\; over\; mathrm\{hypotenuse\; =\; \{3\; over\; sqrt\{x^2\; -\; 8x\; +\; 25,$

and

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

External links

* [http://mss.math.vanderbilt.edu/~pscrooke/MSS/partialfract.html Partial Fraction Expander]
* [http://user.mendelu.cz/marik/maw/index.php?lang=en&form=integral Mathematical Assistant on Web] online calculation of integrals, allows to integrate in small steps (includes partial fractions, powered by Maxima (software))