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Integration Problem

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    How do i integrate:

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

    ?


    2. Relevant equations

    The answer is:

    [tex]\frac{1}{4}(\frac{-2x}{x^2-1}-log(x-1)+log(x+1))[/tex]

    3. The attempt at a solution

    I dont know which method to use.
     
  2. jcsd
  3. Nov 16, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    Re: Integration

    The answer you posted is correct if the integrand is 1/(1-x^2)^2. Is that what you meant? If so, try partial fractions.
     
  4. Nov 16, 2009 #3
    Re: Integration

    Im sorry, the question is correct but the answer is:

    1/2[x/(x^2+1)+arctan(x)]

    here partial fractions wont work because i will still have a term with (x^2+1)^2 in the denominator
     
  5. Nov 16, 2009 #4

    Mark44

    Staff: Mentor

    Re: Integration

    Have you tried a trig substitution? tan (theta) = x would be a good one to start with.
     
  6. Nov 16, 2009 #5

    Dick

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    Science Advisor
    Homework Helper

    Re: Integration

    Ok, then it's a trig substitution. Like x=tan(t).
     
  7. Nov 16, 2009 #6
    Re: Integration

    how did you know what substitution to make??
     
  8. Nov 16, 2009 #7

    Mark44

    Staff: Mentor

    Re: Integration

    Draw a right triangle and label the sides and hypotenuse according to the expression in your integral. Since you have x^2 + 1, that corresponds to the hypotenuse. Label the side opposite your acute angle as x, label the base as 1, and label the hypotenuse as sqrt(x^2 + 1). So tan(theta) = x/1, and sec^2(theta)d(theta) = dx. Use these two equations to completely convert you integral from expressions in x and dx to ones in theta and d(theta).
     
  9. Nov 16, 2009 #8

    Mark44

    Staff: Mentor

    Re: Integration

    If you're asking how did we know to make a trig substitution, they are good bets when you have factors involving the sum or difference of squares, such as sqrt(x^2 + a^2), sqrt(a^2 - x^2), or sqrt(x^2 - a^2).
     
  10. Nov 16, 2009 #9
    Re: Integration

    what if we have (5+x^2)^2 in the denominator?
     
  11. Nov 16, 2009 #10

    Mark44

    Staff: Mentor

    Re: Integration

    Label the opposite side as x and the base as sqrt(x), so tan(theta) = x/sqrt(5)
     
  12. Nov 16, 2009 #11
    Re: Integration

    thank you
     
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