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Homework Help: 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
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 25, 2011 #2
    Hi jessjolt! :smile:

    Firstly, you should know that


    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


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


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


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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


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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:

    The third term can he handled similarly.​

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