You are Here: Home >< Maths

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

Announcements Posted on
TSR's new app is coming! Sign up here to try it first >> 17-10-2016
Thankyou!
2. Partial fractions?

3. (Original post by G07153)
Thankyou!
Expand that brackets first to get x^3 + x

Then write it as (x^3 + x)^-1

Then intergrate as normal
4. (Original post by G07153)
Thankyou!
If you mean 1/(x(x^2 + 1))dx, then that's the same as (x^3 + x)^-1 dx.

What happens when you integrate something to the power of minus 1?

Or make partial fractions, of course
5. I need to use the partial fractions method by the looks of the solutions, but haven't used partial fractions for SO long, could someone run through the method? Thanks for all your help!
6. (Original post by Mr M)
Partial fractions?

How do I go about finding A, B & C??
7. (Original post by G07153)
How do I go about finding A, B & C??
Multiply through by x and then set x = 0. You now have A.

Pick two other values for x (e.g. x = 1 and x = 2) and substitute these to get a pair of simultaneous equations for B and C.
8. (Original post by TGH1)
What happens when you integrate something to the power of minus 1?
You get Zorro.
Spoiler:
Show

Wait.. was is zero?
9. (Original post by fletchdd02)
Then write it as (x^3 + x)^-1

Then intergrate as normal
(Original post by TGH1)
If you mean 1/(x(x^2 + 1))dx, then that's the same as (x^3 + x)^-1 dx.

What happens when you integrate something to the power of minus 1?
I hope you two aren't suggesting what I think you're suggesting... there are two separate reasons why this doesn't work - do you know what they are?
10. (Original post by nuodai)
I hope you two aren't suggesting what I think you're suggesting... there are two separate reasons why this doesn't work - do you know what they are?
what are they?
11. (Original post by G07153)
How do I go about finding A, B & C??
Turn the right hand side into a single fraction over the common denominator x(1+x^2). You do this by cross multiplication.

Therefore when you equate numerators on each side you would have.

1 = (A+B)x^2 + Cx + A

Now compare the coefficients of the powers of x on each side. There is no x^2 term on the left hand side so A+B = 0. Similarly there is no x, so C=0. Finally comparing coefficients of the constants you have 1=A => A=1.

Therefore A+B=0 => 1+B=0 => B=-1.

So the the fraction in partials is 1/x - x/(1+x^2), which you can easily integrate. The numerators are the derivatives (up to a constant) of the denominators.
Remember integral (f'(x)/f(x)) = ln|f(x)|.
12. (Original post by fletchdd02)
what are they?
Maybe just one reason, though it depends on what method you intended. The big one is this: whilst it is the case that (chain rule) it is not the case that .

My guess is that you were advising the OP to do . This isn't the case (differentiate the RHS to check).

This is only the case when is a linear polynomial; for instance it is true that .

In general if you want to reverse the chain rule then you need to use integration by substitution; you can't just divide by the derivative.

The second reason is to do with the way you phrased your post: I was worried you were suggesting changing the power to -2 and dividing or whatever. This is wrong because with integration the power is meant to increase, e.g. . But you can't do this with -1 because then you get 0 and you can't divide by 0. I'm not sure if this is actually what you had in your head, but I thought I'd mention it for completeness.
13. (Original post by nuodai)
...
I'm fairly sure they were suggesting add one to the power and divide by the new power. You just saved them from an abominable and certain death.
14. (Original post by G07153)
Thankyou!
(x power 2 +1) integrate 1/x + whole integrate(derivative of (x power 2 +1) integration of 1/x
now
(x power 2 +1)(natural log of x) + whole integration (2x mutiply natural log of x)
repeat again you will get the solution.
15. (Original post by mahnoorbloch)
(x power 2 +1) integrate 1/x + whole integrate(derivative of (x power 2 +1) integration of 1/x
now
(x power 2 +1)(natural log of x) + whole integration (2x mutiply natural log of x)
repeat again you will get the solution.
You are going to be a barrel of laughs. The spam advertising doesn't help by the way.
16. (Original post by Mr M)
I'm fairly sure they were suggesting add one to the power and divide by the new power. You just saved them from an abominable and certain death.
Ahahaha
17. (Original post by fletchdd02)
Expand that brackets first to get x^3 + x

Then write it as (x^3 + x)^-1

Then intergrate as normal
Would anyone mind doing a step by step method to show how you would integrate this? Sorry but I don't really understand some of the methods posted, I've just finished Core 1 and Core 2 in OCR (I don't recall that a/x method or partial fractions etc in my lessons & they're not in any of my books). I would have originally done what fletchdd did but upon realising that you get zero [and of course you can't divide by zero!] i was shocked!
18. (Original post by sach21sk)
Would anyone mind doing a step by step method to show how you would integrate this? Sorry but I don't really understand some of the methods posted, I've just finished Core 1 and Core 2 in OCR (I don't recall that a/x method or partial fractions etc in my lessons & they're not in any of my books). I would have originally done what fletchdd did but upon realising that you get zero [and of course you can't divide by zero!] i was shocked!
This is A2 maths not AS.
19. (Original post by Mr M)
This is A2 maths not AS.
Haha, phew
20. (Original post by sach21sk)
Haha, phew
Actually, from the thread title, it is undergraduate maths (although most A2 students would probably know how to approach it if prompted).

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 23, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams