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∫(4x - 3) / (x^2 +1) dx

  1. Jun 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Can someone help me integrate (4x - 3) / (x^2 +1) ?

    3. The attempt at a solution

    I'm doing my maths baccalaureate in 2 days and came across this question today!

    I don't know where to start i.e. which method to use:

    Integration by parts definitely does not work, nor does by substitution!

    Thank you!
  2. jcsd
  3. Jun 1, 2010 #2


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    Hi jasper10! :smile:

    Trig subsitution should do it.

    (alternatively, split it into the 4x part and the 3 part … you should be able to integrate those anyway :wink:)
  4. Jun 1, 2010 #3
    easy to do if you use integration by parts since the derivative of the numerator is a constant :D
  5. Jun 1, 2010 #4


    Staff: Mentor

    I think that tiny-tim's advice of splitting the integral into two parts is much simpler than integration by parts - hence, less opportunity for getting things fouled up.
  6. Jun 1, 2010 #5
    Unless I'm sorely mistaken integration by parts is one of the key things the maths part of an IB is supposed to teach you. I think my hint was sufficient and that if appropriate care is taken nothing should get 'fouled-up'.
  7. Jun 1, 2010 #6
    Ok so i get:

    ∫4x / (x^2 + 1) dx - ∫3/ (x^2 + 1) dx

    = 2ln(x^2 + 1) - ???

    How would you integrate: 3 / (x^2 + 1)?

    the annoying part is the "x^2"


    ps: what do you mean by "trig" substitution?

    (I have a feeling this type of question isn't on the syllabus, but i found it on a past back from 2002 - strange)
  8. Jun 1, 2010 #7
    Integration by substitution surely cannot work!

    it's a vicious circle and keeps getting more complicated whilst integrating
  9. Jun 1, 2010 #8


    Staff: Mentor

    I'm not disputing the importance of integration by parts. What I'm saying is that it's always a good idea to use the simplest technique that seems likely to work. For this problem, integration by parts ain't it.
  10. Jun 1, 2010 #9
    Yeah just had another look at it and realised the complication, my apologies.

    To quote myself, apparently I was "sorely mistaken" :S
  11. Jun 1, 2010 #10


    Staff: Mentor

    Apparently you mean "integration by parts."
  12. Jun 1, 2010 #11
    I don't know, nothing seems to work

    I'm off to bed now, maybe tomorrow i can think more clearly.
  13. Jun 1, 2010 #12


    Staff: Mentor

    [tex]\int \frac{dx}{1 + x^2} = tan^{-1}x + C[/tex]

    Presumably you've run across this antiderivative formula and variations of it.
  14. Jun 2, 2010 #13


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    good morning! :smile:
    "trig" substitution is the unofficial but widespread abbreviation for substitution with a trigonometric or hypertrigonometric function (in this case, x = tanu or x = sinhu) :smile:

    but you should have known this one anyway, for the exam …

    check the PF list of standard integrals and memorise a few :wink:

    good luck in your exams! :smile:
  15. Jun 2, 2010 #14

    European baccalaureate or American baccalaureate??? Never heard about the term before ?
  16. Jun 2, 2010 #15
    European baccalaureate..why? :D
  17. Jun 2, 2010 #16
    Ok, thanks Mark44 and tiny-tim!
  18. Jun 2, 2010 #17
    Hadn't heard the term before so I googled it. You can check your abilities to integrate

    with the Wolfram integrator!
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