Partial fractions in integration

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))


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