You are Here: Home >< Maths

# integration help watch

Announcements
1. hi guys, im quite stuck with this question. please give me some hints.
integrate[((x^2)-(1/x))^4]

i've tried to let u=the whole thing inside, but unfortunately i don't know how it works.
thanks
2. I think I'd just multiply it out and then integrate.
3. by binomial expansion? i thought of that before but i think it would be quite time consuming...
4. (Original post by kingsclub)
by binomial expansion? i thought of that before but i think it would be quite time consuming...
It really doesn't take long if you do it instead of thinking about doing it.
5. (Original post by Get me off the £\?%!^@ computer)
It really doesn't take long if you do it instead of thinking about doing it.
but will it work using the substitution method? cause i want to try that method.
6. bump
7. binomial expansion is the best way to go in this case!
8. I just did it, all the x^2 and 1/x are different functions, both being in the bracket of the power of four.
The way I thought of it was to say what 1/x is the same as, as a simpler indice, and then applying one of the laws of indices to expand the whole thing.

Alternatively, what might be a better way is use the chain rule. Add one to the power, divide by the (new power times the derivative of the brackets)
9. (Original post by kingsclub)
but will it work using the substitution method? cause i want to try that method.
If you insist on using a substitution.

Put the term in brackets over a common denominator.

It would be useful practise as you'll require polynomial division, use of partial fractions, a non-trivial (but not difficult) change of variable, practise on fractional indices, and will probably be challenged to show that it gives the same result (I've only worked it partially through, but it looks doable.).

Have fun!
10. (Original post by mckinnon94)
IAlternatively, what might be a better way is use the chain rule. Add one to the power, divide by the (new power times the derivative of the brackets)
I'm not totally sure what you mean by the last paragraph, but it is NOT generally true that .

E.g. (obviously).

By the fallacious "rule": .
11. wow, partial fractions !? something beyond my ability then... i stick to the expansion method then. thanks guys
12. (Original post by kingsclub)
wow, partial fractions
Only probably, I didn't work through all the details.
13. Have you got an answer for it? I think I have a solution but not sure if it's right?
14. (Original post by elementsofsuccess)
Have you got an answer for it? I think I have a solution but not sure if it's right?
http://integrals.wolfram.com/index.j...4&random=false

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:
Updated: April 7, 2011
Today on TSR

### Uni league tables

Do they actually matter?

### University open days

• University of Warwick
Sat, 20 Oct '18
• University of Sheffield
Sat, 20 Oct '18
• Edge Hill University
Faculty of Health and Social Care Undergraduate
Sat, 20 Oct '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams