• February 10th 2010, 02:04 AM
joey1
resolve into partial fraction

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

key ^-to a power eg X^4 is x to a power 4

sorry for the way the question is phrased but i dont know how 2 put x to the power 4 or any integer in any other way ,please could someone demonstrate

regards
jovan
• February 10th 2010, 02:18 AM
HallsofIvy
Quote:

Originally Posted by joey1
resolve into partial fraction

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

key ^-to a power eg X^4 is x to a power 4

sorry for the way the question is phrased but i dont know how 2 put x to the power 4 or any integer in any other way ,please could someone demonstrate

regards
jovan

Hopefully, you know that $x^2- 1= (x- 1)(x+ 1)$ while $x^2+ 1$ cannot be factored with real coefficients. You want to find numbers A, B, C, and D such that
$\frac{x^4}{(x^2+1)(x-1)(x+1)}= \frac{Ax+ B}{x^2+ 1}+ \frac{C}{x- 1}+ \frac{D}{x+ 1}$

Start by multiplying both sides of the equation by $(x^2+ 1)(x- 1)(x+ 1)$ to get rid of the fractions. There are a number of ways to find A, B, C, and D now. The simplest is choosing 4 different values of x to get 4 equations.
• February 10th 2010, 02:31 AM
joey1
thanks hallsofivy
thats a great help
1 more question how did you type http://www.mathhelpforum.com/math-he...64e53b1c-1.gif is there a tool that you used
i would like to learn how to add the power to x instead of typing x^2,it becomes confusing to read sometimes

regards
jovan
• February 10th 2010, 02:49 AM
joey1
thanks hallsofivy
thats a great help
1 more question how did you type http://www.mathhelpforum.com/math-he...64e53b1c-1.gif is there a tool that you used
i would like to learn how to add the power to x instead of typing x^2,it becomes confusing to read sometimes

regards
jovan
• February 10th 2010, 05:06 AM
HallsofIvy
This board is enabled with "LaTex" a very powerful formula formatting language. You access it by typing html code [math ] (without the space) to begin and [/math ] (again without the space) to end and putting LaTex code in between. There is a LaTex tutorial at
http://www.mathhelpforum.com/math-help/latex-help/

You get $x^2+ 1$ simply by typing [math ]x^2+ 1[/math ] (again, without the spaces inside [ ]. If I didn't include those spaces you wouldn't see the code!)

You can do really complicated things such as
$\int_{-\infty}^\infty e^{-x^2} dx= \frac{1}{\sqrt{2\pi}}$

with the code
[math ]\int_{-\infty}^\infty e^{-x^2} dx= \frac{1}{\sqrt{2\pi}}[/math ]

Most of the codes begin with "\". Be sure to distinguish between that and the "/" that ends the LaTex.

"{" and "}" group entries. For example "e^-x^2" without the { } would give
$e^-x^2$.

Finally, you can see the code used for any formula by clicking on that formula.

By the way, while this board uses a version that take [math ] and [/math ], other websites may use [Latex] [/Latex] or .
• February 10th 2010, 09:32 AM
Soroban
Hello, joey1

Quote:

Resolve into partial fractions: . $
\frac{x^4}{(x^2+1)(x^2 - 1)}$

You already noted that the fraction is improper.

It must be divided and then we will deal with the remainder.

So we have: . $\frac{x^4}{x^4-1} \:=\:1 + \frac{1}{x^4-1}$

And we work with the fraction:

. . $\frac{1}{(x-1)(x+1)(x^2+1)} \;=\;\frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}$

We get: . $A = \frac{1}{4},\;\; B = -\frac{1}{4},\;\; C= 0,\;\; D = -\frac{1}{2}$

Answer: . $1 + \frac{\frac{1}{4}}{x-1} - \frac{\frac{1}{4}}{x+1} - \frac{\frac{1}{2}}{x^2+1}$

• February 23rd 2010, 09:39 PM
lebod
solve for THETA
2tan^2theta +sec theta -4=0
• March 1st 2010, 09:06 AM
andrew88
full working of problem
thanx for this solution, but is there someplace I couold be able to see the full working of this problem, im really struggling.