1. ## Partial fractions integrate

integrate -2x+4/(x^2+1)(x-1)^2 dx
How do I start the partial fraction?

ax+b/(x^2 +1) +c/(x-1)^2 +d/(x-1) ??

2. Nothin' to it but to do it. Where did you hide your algebra skills?

Work on your notation skills, too. ax+b/(x^2 + 1) is no good. Think about the order of operations.

I get a = 2 and d = -2

Let's see what you get.

integrate -2x+4/(x^2+1)(x-1)^2 dx
How do I start the partial fraction?

ax+b/(x^2 +1) +c/(x-1)^2 +d/(x-1) ??
Yeah, that's good. Next get the same denominator which can cancel out.

$\displaystyle (ax+b)(x-1)^2 + c(x^2+1) + d(x-1)(x^2+1) = -2x+4$

$\displaystyle f(1): 2c = 2 \: \rightarrow c = 1$

Now we must compare coefficients

$\displaystyle x^3: a+d = 0$

$\displaystyle x^1: a -d = -2$

$\displaystyle x^0: b -d = 3$

Solving the top two equations simultaneously will give values of a and d. You can then use the value of d to find b

4. Originally Posted by TKHunny
Nothin' to it but to do it. Where did you hide your algebra skills?
ax+b/(x^2 + 1) is no good. Think about the .
why is this no good?

why is this no good?
$\displaystyle ax+b/(x^2+1) = ax + \frac{b}{x^2+1}$

Instead you should put a set of brackets around $\displaystyle ax+b$ to give (ax+b)/(x^2+1).

Of course LaTeX is better since $\displaystyle \frac{ax+b}{x^2+1}$ is unambiguous

6. Originally Posted by e^(i*pi)
Yeah, that's good. Next get the same denominator which can cancel out.

$\displaystyle (ax+b)(x-1)^2 + c(x^2+1) + d(x-1)(x^2+1) = -2x+4$

$\displaystyle f(1): 2c = 2 \: \rightarrow c = 1$

Now we must compare coefficients

$\displaystyle x^3: a+d = 0$

$\displaystyle x^1: a -d = -2$

$\displaystyle x^0: b -d = 3$

Solving the top two equations simultaneously will give values of a and d. You can then use the value of d to find b
Thank you ,But I have never done it by comparing coefficients.. can it no be done another way..
I got ax+b(x-1)(x-1)^2 + c(x^2+1)(x-1)^2 +d(x^2+1)(x-1)
===> ax+b(x^3 -3x^2+3x-1) +C(x^3-3x^2+3x-1) +D(x^3-x^2+x-1)=-2x+4
can it not be solved like this by usually putting in values for x?

7. Comparing coefficients is the reliable way to go. the substitution method you suggest has three drawbacks.

1) Emotional turmoil. It causes me pain to substitute values I know are not in the Domain. Nevertheless, it works for lots of things, however...

2) It takes Complex numbers for that x^2 + 1 part. That's likely to be worse than it's worth. Essentially, it doesn't work with strictly quadratic (or higher order) factors.

3) Actually, it doesn't work for ANY quadratic factors. You see that x = 1 is your ONLY substitution. This will not buy you FOUR parameters.

So, if you've ALWAYS only linear factors, and you really don't mind deliberately substituting values not in the Domain, just go ahead and substitute values that will make the individual demoninators zero.

In other words, compare the coefficients. The odd substitution method is a trick that looks compelling in very specific circumstances. Get better at basic algebra (order of operations, for example) and it will not be as bad as you might be conjuring.

8. anyone kind enough to show me please

9. It's just an algebra problem. Get common denominators. Expand, collect, and simplify numerators. Equate coefficients on ALL matching powers of x. e^i*pi has already done most of it. Go look at that post again.

10. Originally Posted by TKHunny
It's just an algebra problem. Get common denominators. Expand, collect, and simplify numerators. Equate coefficients on ALL matching powers of x. e^i*pi has already done most of it. Go look at that post again.
If I thought it was that easy I would do it.

11. Originally Posted by e^(i*pi)

Now we must compare coefficients

$\displaystyle x^3: a+d = 0$

$\displaystyle x^1: a -d = -2$

$\displaystyle x^0: b -d = 3$

Solving the top two equations simultaneously will give values of a and d. You can then use the value of d to find b
Here's the start,

$\displaystyle x^1: a -d = -2\implies a = d-2$

into $\displaystyle x^3: a+d = 0 \implies d-2+d = 0 \implies 2d-2 = 0$

Now what is $\displaystyle d$ ?

If I thought it was that easy I would do it.
Well, there you go. It's a thinking adjustment that is most necessary.

It IS that easy. Follow the steps I listed. Look at the two examples you have been provided. You'll get it. One day, someone will think you make it look easy.

13. Originally Posted by TKHunny
Well, there you go. It's a thinking adjustment that is most necessary.

It IS that easy. Follow the steps I listed. Look at the two examples you have been provided. You'll get it. One day, someone will think you make it look easy.
I dont know why you post here?
Out of all the time you spend talking crap you could be actually helping people. not assuming they should think its easy
Did you ever hear of people learning ?

14. Originally Posted by Bushy
Here's the start,

$\displaystyle x^1: a -d = -2\implies a = d-2$

into $\displaystyle x^3: a+d = 0 \implies d-2+d = 0 \implies 2d-2 = 0$

Now what is $\displaystyle d$ ?
as I said I havent ever arranged coefficients, I dont understand the order in which they are collected.

forget it.

I dont know why you post here?
Out of all the time you spend talking crap you could be actually helping people. not assuming they should think its easy
Did you ever hear of people learning ?
See, it is a thinking problem. I post here to encourage learning. Fortunately, I have infinite patience. I can wait until you stop cussing at me and start paying attention to the many examples you have been given. If you keep it up the way you are going, eventually someone will just work the problem for you (maybe even me) and you will learn less than if you had struggled through the thinking process and developed your own more positive and constructive attitute toward learning.

Here's the very simple idea.

One polynomial: ax^2 + bx + c

Another polynomial: dx^2 + ex + f

These two polynomials are EXACTLY the same if and only if all the corresponding coefficients are identical. This gives:

a = d -- Comparing the x^2 coefficients

b = e -- Comparing the x coefficients

c = f -- Comparing the constant terms.

The idea is very simple. There is no question that the algebra can be a bit messy. You must pay attention and get better at it.

Page 1 of 2 12 Last