Saturday

February 28, 2015

February 28, 2015

Posted by **Chris** on Thursday, April 4, 2013 at 11:24am.

1)x^2/((x-1)(x^2+5x+4))

2)(3x^3-5x^2+12x+4)/(x^4-16)

3)1/(x^2 (x+1)^2 )

4)(x+1)/((x^2+1) 〖(x-1)〗^2

- EDIT -
**EDIT**, Thursday, April 4, 2013 at 11:29am1) x^2

---------------

(x-1)(x^2+5x+4)

2) 3x^3-5x^2+12x+4

-----------------

x^4-16

3) 1

----------

x^2(x+1)^2

4) x+1

--------------

(x^2+1)(x-1)^2

- Math- Partial Fractions -
**Reiny**, Thursday, April 4, 2013 at 12:07pmI will do the first:

let

x^2/((x-1)(x+1)(x+5) = (A/(x-1) + B/(x+1) + C/(x+5)]/((x+1)(x-1)(x+5))

then obviously the common denominator is (x-1)(x+1)(x+5)

then only using the numerator:

A(x+1)(x+5) + B(x-1)(x+5) + C(x-1)(x+1) = x^2

let x = 1 --> 12A = 1 or A = 1/12

let x = 1 --> -8B=1 , or B = -1/8

let x = -5 --> 24C = 25 , or C = 24/25

then x^2/((x-1)(x^2 + 5x + 4))

= 1/(12(x-1)) - 1/((8(x+1)) + 25/(24(x+5))

- Math- Partial Fractions -
**Count Iblis**, Thursday, April 4, 2013 at 1:11pm3) is perhaps more interesting to do, as it's already factored you can just go ahead and solve this in the usual way. But there exist a much faster way to compute partial fraction expansions, which generalizes the method used by Reiny where you insert for x the critical values to simplify the equations.

When you have higher powers, you can in principle still use this method, by taking derivatives of both sides. But that's a bit cumbersome. What works even better is to expand the fraction around each critical point and keep only the singular terms. This is then the desired partial fraction expansion, because the difference between the function and the sum of all the expansions doesn't have any singularities anymore, it must therefore be a polynomial. However, it tends to zero at infinity, and must thus be equal to zero. This argument requires that you consider all the singularities jn the complex plane and that the numerator is of lower degree than the denominator.

The series expansion of

f(x) = 1/(x^2 (x+1)^2 )

around x = 0 is:

1/x^2 *(1 - 2 x + ...) =

1/x^2 - 2/x + non-singular terms.

The series expansion of

1/(x^2 (x+1)^2 )

around x = -1: Put x = -1+t:

1/t^2 *1/(1-t)^2 = 1/t^2 +2/t + non-singular terms =

1/(x+1)^2 + 2/(x+1) + non-singular terms.

The partial fraction expansion is thus:

1/x^2 - 2/x + 1/(x+1)^2 + 2/(x+1)

**Answer this Question**

**Related Questions**

Calc II - Perform long division on the integrand, write the proper fraction as a...

calculus - Decompose 58-x/x^2-6x-16 into partial fractions.

Algebra-Partial Fractions - Can you please help me with the following questions ...

Partial Decompostion Fractions - Can you please help me with the following ...

Binomial - Help me on this one :( Express y= (7-3x-x^2)/[((1-x)^2)(2+x)] in ...

Calc easy - Having trouble getting the correct solution. The integral of “x ...

Calculus - Partial Fractions - What is the integral of 7e^(7t) Divided By e^14t+...

Math - Calculate the integrals by partial fractions and using the indicated ...

Pre-algebra - I dont understand this. Can you do some examples? One student had ...

Maths - If z=(x^n) f(x/y), where f is an arbitrary function, show that: x(...