Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
     \int^1_0 \frac{\ln (\frac{x^2+3x+2}{(x+3)^{2}})}{(x  +3)^{2}}\ dx

    edit: nevermind i see what i did wrong now
    Offline

    3
    ReputationRep:
    (Original post by CancerousProblem)
    Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 1598x154
    <span style="font-family: verdana"><font color="#444444">\int^1_0 </font></span><span style="font-family: verdana"><font color="#444444">\frac{</font></span><span style="font-family: verdana"><font color="#444444">\ln </font></span><span style="font-family: verdana"><font color="#444444">\frac{x+1}{x+3}</font></span><span style="font-family: verdana"><font color="#444444">}{(x+3)^{2}}</font></span><span style="font-family: verdana"><font color="#444444">\</font></span>


     \int^1_0 \frac{\ln \frac{x+1}{x+3}}{(x+3)^{2}}\ dx
    I got 19/12ln3 - 7/3ln2 - 1/6
    Answers got -1/12ln2 less than me
    We can't see the image.
    Offline

    19
    ReputationRep:
    rage inducing member?

    Name:  4.jpg
Views: 71
Size:  8.6 KB
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by lizard54142)
    We can't see the image.
    never added an image, my latex just got bugged
    Offline

    3
    ReputationRep:
    (Original post by CancerousProblem)
    never added an image, my latex just got bugged
    - Latex bugged
    - wrong question
    - wrote out your working wrong
    - incomprehensible substitution

    I am completely lost!
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by lizard54142)
    - Latex bugged
    - wrong question
    - wrote out your working wrong
    - incomprehensible substitution

    I am completely lost!
    lol **** i am so sorry

    the question is correct now
    x+1/x+3 = x was the substitution I used to get the integral
    Offline

    3
    ReputationRep:
    (Original post by CancerousProblem)
    lol **** i am so sorry

    the question is correct now
    x+1/x+3 = x was the substitution I used to get the integral
    Haha no problem, so:

     u = x + \dfrac{1}{x+3} is the substitution?
    Or:  x = u + \dfrac{1}{u+3}?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by lizard54142)
    Haha no problem, so:

     u = x + \dfrac{1}{x+3} is the substitution?
    Or:  x = u + \dfrac{1}{u+3}?
    um no
    (x+1)/(x+3) = u
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by lizard54142)
    Haha no problem, so:

     u = x + \dfrac{1}{x+3} is the substitution?
    Or:  x = u + \dfrac{1}{u+3}?
    nevermind I see where I messed up now thanks. its been a fun 30 minutes with latex rofl
    I should just take a picture next time
    Offline

    3
    ReputationRep:
    (Original post by CancerousProblem)
    nevermind I see where I messed up now thanks. its been a fun 30 minutes with latex rofl
    I should just take a picture next time
    Gosh I was close to doing it as well... where did you go wrong?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by lizard54142)
    Gosh I was close to doing it as well... where did you go wrong?
    I subsituted (x+2)/(x+3) as (x+1)/(x+3) + 1 rofl

    I thought the answers had the same thing but that was before that step
    Offline

    3
    ReputationRep:
    (Original post by CancerousProblem)
    I subsituted (x+2)/(x+3) as (x+1)/(x+3) + 1 rofl

    I thought the answers had the same thing but that was before that step
    That's where I got up to when you said you'd done it haha:

    ln(\dfrac{x+2}{x+3}) = ln(\dfrac{1}{x+3} + \dfrac{x+1}{x+3})
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 14, 2015

1,252

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
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.