NationMaster - Encyclopedia: 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. This article deals with the concept of an integral in calculus. ... In algebra, the partial fraction decomposition of a rational function expresses the function as a sum of fractions, in each term of which, the denominator is an irreducible (i. ... In mathematics, a rational function in algebra is a function defined as a ratio of polynomials. ... In mathematics, a polynomial is an expression in which constants and variables are combined using only addition, subtraction, multiplication, and positive whole number exponents (raising to a power). ...

For an account of how to find this partial fraction expansion of a rational function, see partial fraction. In algebra, the partial fraction decomposition of a rational function expresses the function as a sum of fractions, in each term of which, the denominator is an irreducible (i. ...

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

1 A 1st-degree polynomial in the denominator
2 A repeated 1st-degree polynomial in the denominator
3 An irreducible 2nd-degree polynomial in the denominator
4 A repeated irreducible 2nd-degree polynomial in the denominator
5 External link

A 1st-degree polynomial in the denominator

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

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

$int {1 over u},{du over a}={1 over a}int{duover u}={1 over a}lnleft|uright|+C = {1 over a} lnleft|ax+bright|+C.$

A repeated 1st-degree polynomial in the denominator

The same substitution reduces such integrals as

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

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

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: In mathematics, a discriminant is an expression which discriminates qualities of algebraic structures. ... The introduction to this article provides insufficient context for those unfamiliar with the subject matter. ...

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

and observe that this sum of two squares can never be 0 while x is a real number. In mathematics, the real numbers are intuitively defined as numbers that are in one-to-one correspondence with the points on an infinite lineâ€”the number line. ...

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:

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

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. In mathematics, the absolute value (or modulus1) of a real number is its numerical value without regard to its sign. ...

Next we must treat the integral

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

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

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

Now the substitution

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

$dw=dx/3,$

gives us

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

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

$u=x^2-8x+25,,$

$du=(2x-8),,dx$

$du/2=(x-4),dx.$

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

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

Then we can use a trigonometric substitution: In mathematics, trigonometric substitution is the substitution of trigonometric functions for other expressions. ...

$tantheta={x-4 over 3},,$

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

$dtantheta=sec^2theta,dtheta={dx over 3}.,$

Then the integral becomes

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

By repeated applications of the half-angle formula In mathematics, trigonometric identities are equations involving trigonometric functions that are true for all values of the occurring variables. ...

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

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). In mathematics, the Pythagorean theorem or Pythagorass theorem is a relation in Euclidean geometry between the three sides of a right triangle. ...

Therefore we have

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

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

and

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

External link

Partial Fraction Expander

Category: Calculus

Results from FactBites:

The Logistic Equation and Integration by Partial Fractions (1125 words)
	The Logistic Equation and Integration by Partial Fractions
	To go further, we need to integrate the function shown on the right hand side, but for this, we will use a TRICK: this trick is called Partial Fractions and consists of unraveling the quotient into its simpler fractional pieces.
	The usefulness of this idea resides in the fact that these simpler fractions are easily integrated.

NationMaster - Encyclopedia: Partial fractions in integration (846 words)
	In integral calculus, the use of partial fractions is required to integrate the general rational function.
	In algebra, the partial fraction decomposition of a rational function expresses the function as a sum of fractions, in each term of which, the denominator is an irreducible (i.
	In algebra, the partial fraction decomposition or (partial fraction expansion) is used to reduce the degree of either the numerator or the denominator of a rational function.

More results at FactBites »

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A 1st-degree polynomial in the denominator GA_googleFillSlot("encyclopedia_square");

A repeated 1st-degree polynomial in the denominator

An irreducible 2nd-degree polynomial in the denominator

A repeated irreducible 2nd-degree polynomial in the denominator

External link

A 1st-degree polynomial in the denominator