Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    How Do you intergrate 10/(3+2x)squared

    i am all confused at the moment as my brain has stopped working. :confused:
    Offline

    0
    ReputationRep:
    (Original post by foreverblue)
    How Do you intergrate 10/(3+2x)squared

    i am all confused at the moment as my brain has stopped working. :confused:
    If you differentiate the bottom bit, you get 2. Using the rule that the integral of f'(x)/f(x)=lnf(x) + c and the top is 5 times f'(x), we have:
    INT 5f'(x)/f(x)=5ln(3 + 2x) +c

    EDIT: This is wrong. Note to self - read the question before answering to prevent looking foolish.
    Offline

    1
    ReputationRep:
    (Original post by foreverblue)
    How Do you intergrate 10/(3+2x)squared

    i am all confused at the moment as my brain has stopped working. :confused:
    it's 10(3+2x)^-2 so integrated that would be -5(3+2x)^-1 as it's a function of a function so you need to take the differential of the bracket into account.

    EDIT: That's -5/(3+2x)
    Offline

    2
    ReputationRep:
    (Original post by meepmeep)
    If you differentiate the bottom bit, you get 2. Using the rule that the integral of f'(x)/f(x)=lnf(x) + c and the top is 5 times f'(x), we have:
    INT 5f'(x)/f(x)=5ln(3 + 2x) +c
    wouldn't the 'squared' make any difference?
    Offline

    1
    ReputationRep:
    (Original post by mockel)
    wouldn't the 'squared' make any difference?
    Yes it does, if it was a ln, then the bottom (3+2x)^2 is differentiated for the top, giving the top as 2(3+2x), which is not what you get. I think he just forgot the square, easy mistake to make, I do it as well and I really hope I won't do it tomorrow
    Offline

    0
    ReputationRep:
    (Original post by mockel)
    wouldn't the 'squared' make any difference?
    Oops. Didn't see the squared. Must read the question....:rolleyes:

    Anyway, if that's the case....

    Use integration by substitution. Let 3+2x=u
    Thus, x=(u-3)/2 so dx/du=1/2 so dx=du/2

    Thus integral of 10/(3+2x)^2dx= integral of 5/u^2 = -5u^-1+C
    =-5/(3+2x) + c
    Offline

    1
    ReputationRep:
    (Original post by meepmeep)
    Oops. Didn't see the squared. Must read the question....:rolleyes:

    Anyway, if that's the case....

    Use integration by substitution. Let 3+2x=u
    Thus, x=(u-3)/2 so dx/du=1/2 so dx=du/2

    Thus integral of 10/(3+2x)^2dx= integral of 5/u^2 = -5u^-1+C
    =-5/(3+2x) + c
    Must read the question is what was hissed at me fiercely before my physics practical exam. Apparently I am god's gift to not reading questions :rolleyes:
    Offline

    0
    ReputationRep:
    (Original post by Mysticmin)
    Must read the question is what was hissed at my fiercely before my physics practical exam. Apparently I am god's gift to not reading questions :rolleyes:
    And I keep doing "Andrew specials" as they're now called (not by me though) which is basically doing all the hard parts right then do something completely stupid at the end (eg. 1/4+4=5/4).


    Awww, just realised, I've gone and missed my thousandth post.
    Offline

    0
    ReputationRep:
    (Original post by Mysticmin)
    it's 10(3+2x)^-2 so integrated that would be -5(3+2x)^-1 as it's a function of a function so you need to take the differential of the bracket into account.

    EDIT: That's -5/(3+2x)
    So when integrating something similar to the question given above, i not only divide by the original power plus one, but also the DIFFERENTIAL of the funtion underneath - is that right?
    Offline

    1
    ReputationRep:
    (Original post by Silly Sally)
    So when integrating something similar to the question given above, i not only divide by the original power plus one, but also the DIFFERENTIAL of the funtion underneath - is that right?
    Correct
    Offline

    1
    ReputationRep:
    (Original post by meepmeep)
    And I keep doing "Andrew specials" as they're now called (not by me though) which is basically doing all the hard parts right then do something completely stupid at the end (eg. 1/4+4=5/4).


    Awww, just realised, I've gone and missed my thousandth post.
    Damn, well have a theoretical chocolate cake complete with a thousand lit candles from me

    I always mess up the easy bits, I switch off when I do them. You're called andrew too? Why is it all the andrews I meet do double maths... Must be the name.
    Offline

    1
    ReputationRep:
    (Original post by foreverblue)
    How Do you intergrate 10/(3+2x)squared

    i am all confused at the moment as my brain has stopped working. :confused:
    ∫ ² ³ ± ÷ ° √ ∂ ∞ ≈ ≠ ≡ ≥ ≤ ø Ø ½ ¼ « » µ π

    Or if you want to keep it very simple, take the 10 outside of the integral:

    10∫ 1/(3+2x)^2 dx and integrate:

    10 * -((3+2x)^-1)/2) = -5(3+2x)^-1 + c
    Offline

    0
    ReputationRep:
    (Original post by Mysticmin)
    Correct
    YAY thanks!!!

    So if the power is more than or less than one, i basically put it on the top line and make it a -ve power and then take it from there.

    Also one other question, suppose if the question was something like:

    INT 10/[(2x^2 - 3x +1)^2]

    Would the answer be:

    [-10(4x - 3)]/[(2x^2 - 3x +1] ????

    This is a made up question that i thought of btw
    Offline

    1
    ReputationRep:
    (Original post by Silly Sally)
    YAY thanks!!!

    So if the power is more than or less than one, i basically put it on the top line and make it a -ve power and then take it from there.

    Also one other question, suppose if the question was something like:

    INT 10/[(2x^2 - 3x +1)^2]

    Would the answer be:

    [-10(4x - 3)]/[(2x^2 - 3x +1] ????

    This is a made up question that i thought of btw
    No, as the x's on top can't dissapear. I'm sorry, You can only do that when the bracket differentiates to give an integer

    With your above question, you'd need to partial fraction the function before integrating.
    Offline

    2
    ReputationRep:
    (Original post by Mysticmin)
    No, as the x's on top can't dissapear. I'm sorry, You can only do that when the bracket differentiates to give an integer

    With your above question, you'd need to partial fraction the function before integrating.
    cant you expand it and then 10 divided by the coefficients of the equation then differentiate it :confused:
    Offline

    1
    ReputationRep:
    (Original post by Mysticmin)
    No, as the x's on top can't dissapear. I'm sorry, You can only do that when the bracket differentiates to give an integer

    With your above question, you'd need to partial fraction the function before integrating.
    Oh dear, I wonder how silly sally feels. Who ever said maths was easy? Why can't we get nice questions. Oh well.
    Offline

    0
    ReputationRep:
    (Original post by Mysticmin)
    No, as the x's on top can't dissapear.
    What do you mean by this statement?

    Ok - i PROMISE the last question!!!

    suppose if you have an x at the top before you integrate so for example - here's another of my made up examples!!! :

    INT x/(10-3x)^3

    would the answer to my made up question be:

    (x/6)/(10 - 3x)^2 ???

    Or wouldn't it work?

    Thanks!!!

    I will give you a rep for all your help - but you will have to wait for a while until the 24hr rule has expired!!!
    Offline

    0
    ReputationRep:
    (Original post by Bhaal85)
    Oh dear, I wonder how silly sally feels. .
    Awful? Terrible? Nearly close to tears? Suicidal?


    Oh NO!!! - Those are just an understatement at the moment!!!
    Offline

    0
    ReputationRep:
    (Original post by Silly Sally)
    What do you mean by this statement?

    Ok - i PROMISE the last question!!!

    suppose if you have an x at the top before you integrate so for example - here's another of my made up examples!!! :

    INT x/(10-3x)^3

    would the answer to my made up question be:

    (x/6)/(10 - 3x)^2 ???

    Or wouldn't it work?

    Thanks!!!
    Anyone???
 
 
 
Turn on thread page Beta
Updated: June 8, 2004

5,362

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.