∫(x-1∕x)² dx Help, please

Announcements Posted on
Four things that unis think matter more than league tables 08-12-2016
    • Thread Starter
    Offline

    2
    ReputationRep:
    The question is in the title.
    So far I have written it as
    ∫x²-(x-1)²dx
    and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
    I've done the first x
    ∫x² dx =x³/3
    Offline

    3
    ReputationRep:
    (Original post by TheGreatPumpkin)
    The question is in the title.
    So far I have written it as
    ∫x²-(x-1)²dx
    and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
    I've done the first x
    ∫x² dx =x³/3
    Expand (x- \frac{1}{x})^2 = (x-\frac{1}{x})(x-\frac{1}{x})

    The result of (a+b)^2 is a^2 +2ab +b^2, that is the first term squared, plus two times the product, plus the last term squared.

     \int (x- \frac{1}{x})^2 dx = \int(x^2 - 2 + \frac{1}{x^2})dx

    From here it can be integrated easily.
    Offline

    2
    ReputationRep:
    (Original post by TheGreatPumpkin)
    The question is in the title.
    So far I have written it as
    ∫x²-(x-1)²dx
    and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
    I've done the first x
    ∫x² dx =x³/3
    You can't go from ∫(x-1∕x)² to ∫x^2 - (x^-1)^2. You need to use the reverse chain rule so you first need to differentiate the bracket so you get dy/dx = 1 + x^-2. Using the reverse chain rule you get ((x-1/x)^3)/(3(1+x^-2)). Any math geniuses agree? Quite new to integration so sorry for any mistake. Hope it helps
    Offline

    2
    ReputationRep:
    (Original post by NotNotBatman)
    Expand (x- \frac{1}{x})^2 = (x-\frac{1}{x})(x-\frac{1}{x})

    The result of (a+b)^2 is a^2 +2ab +b^2, that is the first term squared, plus two times the product, plus the last term squared.

     \int (x- \frac{1}{x})^2 dx = \int(x^2 - 2 + \frac{1}{x^2})dx

    From here it can be integrated easily.
    Any help in what I did wrong?
    Offline

    3
    ReputationRep:
    (Original post by samendrag)
    Any help in what I did wrong?
    Yeah, (x-\frac{1}{x})^2 \not= x^2-(x^{-1})^2
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    Yeah, (x-\frac{1}{x})^2 \not= x^2-(x^{-1})^2
    Could you use the reverse chain rule, how would you use it in this question?
    Offline

    3
    ReputationRep:
    (Original post by samendrag)
    You can't go from ∫(x-1∕x)² to ∫x^2 - (x^-1)^2. You need to use the reverse chain rule so you first need to differentiate the bracket so you get dy/dx = 1 + x^-2. Using the reverse chain rule you get ((x-1/x)^3)/(3(1+x^-2)). Any math geniuses agree? Quite new to integration so sorry for any mistake. Hope it helps
    (Original post by samendrag)
    Any help in what I did wrong?
    x is a variable factor, so the reverse chain rule doesn't apply. You can divide by constants when using the reverse chain rule, but not by variable factors.
    Offline

    2
    ReputationRep:
    (Original post by NotNotBatman)
    x is a variable factor, so the reverse chain rule doesn't apply. You can divide by constants when using the reverse chain rule, but not by variable factors.
    ohh ok thanks!
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by NotNotBatman)
    Expand (x- \frac{1}{x})^2 = (x-\frac{1}{x})(x-\frac{1}{x})

    The result of (a+b)^2 is a^2 +2ab +b^2, that is the first term squared, plus two times the product, plus the last term squared.

     \int (x- \frac{1}{x})^2 dx = \int(x^2 - 2 + \frac{1}{x^2})dx

    From here it can be integrated easily.
    Thanks, I forgot how to expand for a moment
    Offline

    3
    ReputationRep:
    (Original post by samendrag)
    Could you use the reverse chain rule, how would you use it in this question?
    Apologies, that reply was meant for OP. The chain rule wouldn't be of much help here.
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  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
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: October 16, 2016
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.

Poll
Do you think you'll achieve your predicted A Level grades?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.