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Integrate using partial fractions (arctan x) / ((x+1)(x^2+1))?

  1. Jun 25, 2011 #1
    Integrate using partial fractions (arctan x) / ((x+1)(x^2+1))?

    My attempt:

    (arctan x) / ((x+1)(x^2+1)) = A/(x+1) + (Bx+C)/(x^2+1)
    arctan x = A(x^2+1) + (Bx+C)(x+1)
    when x= -1, A= -pi/8

    so plugging in -pi/8 for B makes:
    x= tan[ (B-(pi/8))x^2 + (B+C)x + (C-(pi/8)) ]
    and when x=0, C= pi/8

    so plugging in pi/8 for C and rearranging produces:
    x=tan[ B(x^2+x) + (((pi*x)/8)*(1-x)) ]
    and when x=1, B=pi/8

    so (arctan x) / ((x+1)(x^2+1)) = [(-pi/8)/(x+1)] + {{ [(pi*x)/8] + (pi/8) }/(x^2+1) }

    but when i graph in radian mode on a graphic calculator (arctan x) / ((x+1)(x^2+1)) and my result, they are not the same thing, so my answer must be wrong. The graphs appear very similar though, so it seems there is only a small error? Please help?? :D
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    3. The attempt at a solution
     
  2. jcsd
  3. Jun 25, 2011 #2

    micromass

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    Hi jessjolt! :smile:

    Firstly, you should know that

    [tex]\int{\frac{arctan(x)}{(x+1)(x^2+1)}dx}[/tex]

    is not solvable using elementary functions.

    Secondly, what you did is sadly not allowed. You wrote

    [tex]arctan x = A(x^2+1) + (Bx+C)(x+1)[/tex]

    But this would imply that the arctangent was a polynomial, which is not not the case. What you could do is split

    [tex]\frac{1}{(x+1)(x^2+1)}[/tex]

    into partial fractions and then multiply it all by arctan(x). But, then again, the integral isn't solvable...
     
  4. Jun 25, 2011 #3

    eumyang

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    You can't use partial fractions here, because your original function is NOT a rational function. (Rational functions are in the form of f(x)/g(x), where f and g are polynomials.)


    EDIT: Beaten to it! :biggrin:
     
  5. Jun 25, 2011 #4
    oohhh ok i see thanks...so there is no way to use partial fractions with transcendental functions like the arctangent?
     
  6. Jun 25, 2011 #5

    micromass

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    Not that I'm aware of, no.
     
  7. Jun 25, 2011 #6

    SammyS

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    Well you can expand [itex]\frac{1}{(x+1)(x^2+1)}[/itex] using partial fractions, but I'm quite that won't help with this integral.
     
  8. Jun 26, 2011 #7
    you could write the arctan(x) in term of natural logs . and then factor
    [itex] (x^2+1) [/itex] as (x+i)(x-i) but i dont know if that will help .
     
  9. Jun 26, 2011 #8

    Char. Limit

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    I think that if you write tan-1(x) as an infinite series, you can integrate it then. Of course, it will be hell, and your integral will probably be in a series.
     
  10. Jun 26, 2011 #9

    SammyS

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    (OP seems to have dropped out of this discussion.)

    Ooooo ... I like this idea ! However, I wouldn't want to work it all out.

    But, it does lead to the following hybrid scheme:
    Using partial fractions, [tex]\frac{1}{(x+1)(x^2+1)}=\frac{1}{2(x^2+1)}-\frac{x}{2(x^2+1)}+\frac{1}{2 (x+1)}[/tex]
    The first term will give [itex]\displaystyle \int{\frac{\tan^{-1}(x)}{2(x^2+1)}}\,dx\,,[/itex] which is not a difficult integration.

    For the 2nd & third terms, use the infinite series expansion for arctan(x), expanded about x = 0. (It has only odd powers of x.)
    Multiplying this series by the second term results in a series whose terms are of the form:
    [tex]\frac{x^{2n}}{(x^2+1)}=x^{2(n-1)}-x^{2(n-2)}+x^{2(n-3)}-\dots-(-1)^nx^2(-1)^n+\frac{(-1)^n}{x^2+1}[/tex]

    The third term can he handled similarly.​

    Char. Limit is right. This is a mess - but do-able.
     
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