The Student Room Group

Integrating a fraction

How would you integrate

2x^2/(1+x) ?

Attempted to use partial fractions but not successful ?
Reply 1
Original post by jon2016
How would you integrate

2x^2/(1+x) ?

Attempted to use partial fractions but not successful ?


Long division. You should get 2x21+x=2x2x1+x\frac{2x^2 }{1+x} = 2x - \frac{2x}{1+x} then long division on the second fraction as well.
Reply 2
substitution?

u=1+x
(edited 7 years ago)
Reply 3
How exactly would long division work with polynomials ? The X in the numerator cant cancel out ?

Original post by Zacken
Long division. You should get 2x21+x=2x2x1+x\frac{2x^2 }{1+x} = 2x - \frac{2x}{1+x} then long division on the second fraction as well.


The textbook does this http://imgur.com/a/ABxPl

Not sure what method they followed ?
Reply 4
Original post by jon2016
How exactly would long division work with polynomials ? The X in the numerator cant cancel out ?



The textbook does this http://imgur.com/a/ABxPl

Not sure what method they followed ?


What they've done is 2x21+x=2x2+2x2x1+x=2x(x+1)2x1+x=2x(1+x)1+x2x1+x=2x2x1+x\frac{2x^2}{1+x} = \frac{2x^2 + 2x - 2x}{1 +x} = \frac{2x(x+1) - 2x}{1+x} = \frac{2x(1+x)}{1+x} - \frac{2x}{1+x} = 2x - \frac{2x}{1+x}

No perform the same 'trick' or 'technique' on the second fraction by 2x=2x+22=2(1+x)22x = 2x + 2 - 2 = 2(1+x) - 2
Reply 5
Original post by Zacken
What they've done is 2x21+x=2x2+2x2x1+x=2x(x+1)2x1+x=2x(1+x)1+x2x1+x=2x2x1+x\frac{2x^2}{1+x} = \frac{2x^2 + 2x - 2x}{1 +x} = \frac{2x(x+1) - 2x}{1+x} = \frac{2x(1+x)}{1+x} - \frac{2x}{1+x} = 2x - \frac{2x}{1+x}

No perform the same 'trick' or 'technique' on the second fraction by 2x=2x+22=2(1+x)22x = 2x + 2 - 2 = 2(1+x) - 2


wow thats amazing, I would never have thought of doing that in the exam ! thanks so much.

Out of curiosity, are there any clear things to look for , so you know which integration technique to use ? from partial, sub and by parts ?
Reply 6
Original post by jon2016
wow thats amazing, I would never have thought of doing that in the exam ! thanks so much.

Out of curiosity, are there any clear things to look for , so you know which integration technique to use ? from partial, sub and by parts ?


The thing is, here - long division and substitution would have worked as well as just this basic simplification. You might want to learn "long division for polynomials" so you're able to do what I did in my previous post but without it requiring any creative thought and it's a lot more systematic.

You tend to get used to what technique to use once you've done enough practice; normally IBP is when you have a product of two functions where each is easy integrable/differentiable but put together they aren't, etc...
Reply 7
Original post by Zacken
The thing is, here - long division and substitution would have worked as well as just this basic simplification. You might want to learn "long division for polynomials" so you're able to do what I did in my previous post but without it requiring any creative thought and it's a lot more systematic.

You tend to get used to what technique to use once you've done enough practice; normally IBP is when you have a product of two functions where each is easy integrable/differentiable but put together they aren't, etc...

What about partial fractions ?
Could we have used partial fractions for the above question ?
Reply 8
Original post by jon2016
What about partial fractions ?
Could we have used partial fractions for the above question ?


Well, no - you can only use partial fractions when there is more than one factor in the denominator. But all you have here is (1+x)(1+x) in the denominator, which you can't do anything with, unfortunately.
Reply 9
Original post by Zacken
Well, no - you can only use partial fractions when there is more than one factor in the denominator. But all you have here is (1+x)(1+x) in the denominator, which you can't do anything with, unfortunately.


what about e^ root(x) / root(x)

My textbook does it by Integration by substitution, but the fact is has two terms, suggested to me it should be Integration by parts ?
Reply 10
Original post by jon2016
what about e^ root(x) / root(x)

My textbook does it by Integration by substitution, but the fact is has two terms, suggested to me it should be Integration by parts ?


You can do it by parts, if you want:
Unparseable latex formula:

u = e^{\sqrt{x}} \Rightarrow \frac{\mathrn{d}u}{\mathrm{d}x} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}

.

And dv=1xv=2x\mathrm{d}v = \frac{1}{\sqrt{x}} \Rightarrow v= 2\sqrt{x}.

And that would work out fine.
Reply 11
Original post by Zacken
You can do it by parts, if you want:
Unparseable latex formula:

u = e^{\sqrt{x}} \Rightarrow \frac{\mathrn{d}u}{\mathrm{d}x} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}

.

And dv=1xv=2x\mathrm{d}v = \frac{1}{\sqrt{x}} \Rightarrow v= 2\sqrt{x}.

And that would work out fine.


ah right thanks, it didnt occur to me

I thought about what you said about being creative

Is this a legitimate maths rearranging technique http://i.imgur.com/i1kWglN.jpg ie multpiply numerator and denominator by 2
Reply 12
Original post by jon2016


Is this a legitimate maths rearranging technique http://i.imgur.com/i1kWglN.jpg ie multpiply numerator and denominator by 2


Very much so! You learnt about this technique in middle school, I'm sure. How do you add 12+14\frac{1}{2} + \frac{1}{4}. Well, you take the 12\frac{1}{2} and multiplied the numerator and denominator by 22 to get 24\frac{2}{4}. So that you could write 12+14=24+14=34\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}.

So, to answer your question, yes: what you've done is valid.

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