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# Help me integrate this please :) watch

1. (Original post by alpha_4)
Thanks, I don't think those types of partial fractions are in the c4 syllabus but it seems like the most sensible option

Only one more, if you don't mind...

x^2/X^3 + 1
Technically, everything you need to know about partial fractions IS on the C4 syllabus, you just need to be able to recognise patterns.

I.e, if it's an improper fraction, instead of using long division, you can do (using an example)

= .

This make it's much quicker and works for fractions which might look very difficult to use LD on. (I think I've done that right, but I've picked a stupid fraction to try and work out).

Anyway, there's a simple rule I've discovered (I'm pretty sure it's in a textbook somewhere, but it works all the same). If the numerator has a variable with a power of 1, and the denominator has a variable with a power of 1, n-d = 0, which means that one of the factors is just a constant, i.e .

If the numerator has a power of 2 and the denominator has a power of 1, then the partial will include a polynomial of order 1, i.e, .

If the numerator has a power of 3 and the denominator has a power of 1, then the partial will include a polynomial of order 2 i.e, .

Etc ...

You touch upon it a little in C4, but like I said, you need to be able to use your intuition a little. Being able to use partial fractions like that comes in very handy for some C4 questions, and in particularly, M3+ and FP3 integrals.
2. (Original post by TwilightKnight)
Technically, everything you need to know about partial fractions IS on the C4 syllabus, you just need to be able to recognise patterns.

I.e, if it's an improper fraction, instead of using long division, you can do (using an example)

= .

This make it's much quicker and works for fractions which might look very difficult to use LD on. (I think I've done that right, but I've picked a stupid fraction to try and work out).
Technically, what you've done there is long division (to get the constant A + a remainder) and then used partial fractions on the remainder. I'm not too sure what you're getting at, I'd be interested to hear what you were meaning to say.

Anyway, there's a simple rule I've discovered (I'm pretty sure it's in a textbook somewhere, but it works all the same). If the numerator has a coefficient with a power of 1, and the denominator has a coefficient with a power of 1, n-d = 0, which means that one of the factors is just a constant, i.e .
This doesn't make sense, coefficient of what? A coefficient is the scalar multiplier of any term involving a variable - so saying the "coefficient of the numerator is 1" doesn't really have any meaning.

If the numerator has a power of 2 and the denominator has a power of 1, then one of the factors will be a polynomial of order 1, i.e, .

If the numerator has a power of 3 and the denominator has a power of 1, then one of the factors will be a polynomial of order 2 i.e, .
Again, you seem to be explaining long division. It's not a 'factor' that will result from considering these divisions but rather that you would get a quotient (in the forms you have stated) plus a remainder (an proper fraction).

What you've done here is just state what the result of polynomial division is. Partial fractions are usually applied to fractions where the polynomial in the denominator is of greater degree than the polynomial in the numerator.
3. (Original post by Farhan.Hanif93)
.
I don't know why you've decided to shoot me down, but I'll explain:

Partial fractions we break down are the reverse of using either + or - operations on Algebraic fractions. What you're doing when you're using a partial fraction is decomposing it into it's initial fractions that make up the full fraction.

All I was saying, was by using the method I've stated, you can skip directly using long division and just treat the whole thing as you would a normal fraction you were decomposing. Whereas if you were using LD, you would be directly looking for the values of the unknowns, which could potentially be a time waster, and also are very tricky to manipulate in some situations.

In the example I gave, that would mean that after putting it in the terms I gave, you simply multiply through by the denominator on the left hand side, giving .

I wasn't trying to say that I was doing anything groundbreaking, all I was saying that was in order to use the method of partial fraction decomposition that people doing A Level Maths are familiar with, you can treat an improper fraction in the way I said, shortening it to 1 step and making integration etc quicker and cleaner.
4. (Original post by Farhan.Hanif93)
This doesn't make sense, coefficient of what? A coefficient is the scalar multiplier of any term involving a variable - so saying the "coefficient of the numerator is 1" doesn't really have any meaning.
I realise that bit didn't make sense. I was halfway through typing out "after multiplying, you can then compare the coefficients of the powers of 'x' to work out the unknowns", but realised that that may have made things harder to follow, so I deleted it and re-wrote it, but obviously not right.
5. (Original post by TwilightKnight)
I don't know why you've decided to shoot me down, but I'll explain:
Oh no, don't get me wrong, I wasn't aiming to shoot you down. As I said in my post, I was interested to hear what you were getting at.

Partial fractions we break down are the reverse of using either + or - operations on Algebraic fractions. What you're doing when you're using a partial fraction is decomposing it into it's initial fractions that make up the full fraction.

All I was saying, was by using the method I've stated, you can skip directly using long division and just treat the whole thing as you would a normal fraction you were decomposing. Whereas if you were using LD, you would be directly looking for the values of the unknowns, which could potentially be a time waster, and also are very tricky to manipulate in some situations.

In the example I gave, that would mean that after putting it in the terms I gave, you simply multiply through by the denominator on the left hand side, giving .

I wasn't trying to say that I was doing anything groundbreaking, all I was saying that was in order to use the method of partial fraction decomposition that people doing A Level Maths are familiar with, you can treat an improper fraction in the way I said, shortening it to 1 step and making integration etc quicker and cleaner.
That's all fine, I was wondering why you brought it up when the method you just explained was the method that was being encouraged throughout the thread anyway. I thought you were on to something that I had missed in my explanations or something, or a shortcut, so I wanted to know what you meant but that doesn't seem to be the case.
6. (Original post by Farhan.Hanif93)
That's all fine, I was wondering why you brought it up when the method you just explained was the method that was being encouraged throughout the thread anyway. I thought you were on to something that I had missed in my explanations or something, or a shortcut, so I wanted to know what you meant but that doesn't seem to be the case.
Nope. I wasn't trying to do anything special. The OP just said that he hadn't seen partial fractions done like that before, so I was just explaining to him what was going on when they were being done in your examples.
7. (Original post by TwilightKnight)
Nope. I wasn't trying to do anything special. The OP just said that he hadn't seen partial fractions done like that before, so I was just explaining to him what was going on when they were being done in your examples.
Sure, that's fair enough. I'm still not sure whether you had addressed his concerns, though. He appeared to be asking how to deal with fractions of the form:

Where is a polynomial of degree , is a polynomial of degree and .

Which I'm not too sure you explained how to deal with.

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