Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    k;
    • PS Helper
    Offline

    14
    PS Helper
    Well how else can you write \dfrac{u+1}{u}?
    Offline

    2
    ReputationRep:
    (Original post by hisname)
    How would i integrate x/(x-1)?

    I've taken a substitution of u=x-1 and re-written the expression as

    (U+1)/U but i don't know what do next i'm guessing its some sort of natural log but i'm not sure.

    Thanks in advance for your help.

    Spoiler:
    Show
    split the fraction.

    ANS:  u + Ln(u) + C


    _Kar.
    • Thread Starter
    Offline

    0
    ReputationRep:
    ';';
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by nuodai)
    Well how else can you write \dfrac{u+1}{u}?
    How exactly would i split the fraction like this

    x* 1(x-1) and do integration by parts.
    Offline

    11
    ReputationRep:
    (Original post by hisname)
    How exactly would i split the fraction like this

    x* 1(x-1) and do integration by parts.
    You're creating too much work for yourself.

    Think about what fractions adding and subtracting actually does.

    I.e, x/4 + 2y/4 = x+2y/4

    You can then split it up again into the same fractions.

    When two coefficients share a denominator by the action of addition or subtraction, you can split them up simply by putting the sign in the middle and putting them over separate denominators of the same value.

    Think about what you would get in this case.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Is this right? you can re-write (U+1)/U as (U/U)+(1/u) or (1+(1/u))

    On integrating that gives u+ ln u +c
    • PS Helper
    Offline

    14
    PS Helper
    (Original post by hisname)
    Is this right? you can re-write (U+1)/U as (U/U)+(1/u) or (1+(1/u))

    On integrating that gives u+ ln u +c
    Yup, and then you can substitute back to get it in terms of x.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Thank you very much for all your help everyone.
    Offline

    1
    ReputationRep:
    no you need to differentiate u with respect to x because of the product rule, and multiply it by the function where u is substituted for x to find your integrand
    • PS Helper
    Offline

    14
    PS Helper
    (Original post by thievingllama)
    no you need to differentiate u with respect to x because of the product rule, and multiply it by the function where u is substituted for x to find your integrand
    The product rule definitely doesn't come into this, but if you're referring to the part in integration by substitution where you multiply by du/dx, it doesn't matter in this case since it evaluates to 1.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 2, 2011
Poll
Do I go to The Streets tomorrow night?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

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

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

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