Outils :

Vous avez un site web ? Un blog ?

Partial fractions in integration

From Wikipedia, the free encyclopedia

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.

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 External links

[edit] A 1st-degree polynomial in the denominator

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

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

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

[edit] A repeated 1st-degree polynomial in the denominator

The same substitution reduces such integrals as

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

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

[edit] 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

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

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

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

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:

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

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:

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

Now the substitution

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

$dw=dx/3\,$

gives us

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

Putting it all together,

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

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

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

Then we can use a trigonometric substitution:

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

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

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

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

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

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

and

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

[edit] 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))

rencontre

Partial fractions in integration - En savoir plus

Encyclopédie Recherche.fr - Partial fractions in integration

Rencontre Partial fractions in integration - Articles à la une

"Je rencontre quelques peines, je rencontre beaucoup de joie, c'est parfois une question de chance, souvent une rencontre de choix."

© 2009 Netencyclo - Netencyclo Home - Terms of Service - Privacy Policy - Program Policies
Netencyclo, the Wikipedia mirror : the biggest multilingual free-content encyclopedia on the Internet. Cet article, miroir de l'article de Wikipédia est conforme aux termes de la GFDL All Wikipedia content is licensed under the GNU Free Documentation License (see details). Content on this web site is provided for informational purposes only. We accept no responsibility for any loss, injury or inconvenience sustained by any person resulting from information published on this site. We encourage you to verify any critical information with the relevant authorities.

- Partial fractions in integration -