Turn on thread page Beta
    • Thread Starter
    Offline

    16
    ReputationRep:
    Just refreshing my memory on these.

    \int \frac{3}{x} dx = 3lnx + c

    \int \frac{2}{3x} dx = \frac{2}{3}lnx + c

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

    please check (I think the second one is wrong)

    edit: Also

    how would you do \int \frac{3}{5x+1} dx
    Offline

    2
    ReputationRep:
    (Original post by Prokaryotic_crap)
    Just refreshing my memory on these.







    please check (I think the second one is wrong)
    All seem correct to me.
    Offline

    10
    ReputationRep:
    All correct!
    Offline

    2
    ReputationRep:
    I swear, at some point in my C4 exam, I'm going to put my head in my hands, then scream "I don't know who I am any more"

    Thank god I only need a B.
    Offline

    0
    ReputationRep:
    (Original post by Prokaryotic_crap)
    edit: Also

    how would you do \int \frac{3}{5x+1} dx
    Make a substitution u=5x+1.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by IchiCC)
    All seem correct to me.
    am I wrong or whenever you differentiate ln(ax) you always get \frac{1}{x}
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Gemini92)
    Make a substitution u=5x+1.
    oh yeah
    Offline

    0
    ReputationRep:
    (Original post by Prokaryotic_crap)
    am I wrong or whenever you differentiate ln(ax) you always get \frac{1}{x}
    You can rewrite ln(ax) as  lna + lnx and differentiate that. lna is a constant so it differentiates to 0 so you're left with the derivative of lnx which is 1/x.
    Offline

    1
    ReputationRep:
    (Original post by Prokaryotic_crap)
    edit: Also

    how would you do \int \frac{3}{5x+1} dx

    3/5ln(5x+1)

    If the differential of the bottom is on the top, then its just ln of the bottom times by whatever constant you need to get it to work.
    Offline

    2
    ReputationRep:
    (Original post by Gemini92)
    Make a substitution u=5x+1.
    Actually, I don't think you'd need to do that substitution.

    you'd simply get
     5/3 ln(5x +1)

    I think...
    Offline

    0
    ReputationRep:
    (Original post by fibrebiz)
    Actually, I don't think you'd need to do that substitution.

    you'd simply get
     5/3 ln(5x +1)

    I think...
    It'd be 3/5 ln(5x+1)
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Gemini92)
    You can rewrite ln(ax) as  lna + lnx and differentiate that. lna is a constant so it differentiates to 0 so you're left with the derivitive of lnx which is 1/x.
    that makes sense, never thought of that. last night i was differentiating ln2x and done \frac{d}{dx}(ln2x}) = \frac{1}{2x}\times 2 = \frac{1}{x} and then I noticed I always get 1/x regardless of the coefficient, so I thought I might be doing it wrong
    Offline

    12
    ReputationRep:
    (Original post by Prokaryotic_crap)
    edit: Also

    how would you do \int \frac{3}{5x+1} dx
    Making a substitution makes it more complicated than it needs to be. Whenever possible, try to do it by inspection.

    If you consider \dfrac{d}{dx}[\mathrm{ln}(5x+1)], we get

    \dfrac{5}{5x+1}

    This is close to what we want, but not quite, as we have a 3 in the numerator not a 5, so we can adjust the numerical factor to account for this. If we put a factor of \dfrac{3}{5} infront to get,

    \dfrac{d}{dx} [\frac{3}{5}\mathrm{ln}(5x+1)], we get

    \dfrac{3}{5x+1}, hence \int \frac{3}{5x+1} dx = \frac{3}{5}\mathrm{ln}(5x+1) +c
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 26, 2011

University open days

  1. Norwich University of the Arts
    Postgraduate Open Days Postgraduate
    Thu, 19 Jul '18
  2. University of Sunderland
    Postgraduate Open Day Postgraduate
    Thu, 19 Jul '18
  3. Plymouth College of Art
    All MA Programmes Postgraduate
    Thu, 19 Jul '18
Poll
How are you feeling in the run-up to Results Day 2018?
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.