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Integral of (x-1)/(x^2-4x+5)

  1. Mar 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the integral of (x-1)/(x^2-4x+5)

    2. Relevant equations

    3. The attempt at a solution

    After completing the square I get
    Using substitutions x-2=tanθ; x-1=tanθ+1; dx=sec^2θ*dθ
    After cancelling
    integrating this I get
    After substituting x back in
    My answer key has a 1/2 infront of the natural log and I'm at a loss as to how they get it.
  2. jcsd
  3. Mar 24, 2015 #2


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    I think if you do the substitution properly, you will see that x^2-4x+5 = sec^2(θ), not sec(θ).
  4. Mar 24, 2015 #3
    I'm confused because with my substitution x^2-4x+5 turns into sec^2(θ) after using the identity tan^2(θ)+1=sec^2(θ)
  5. Mar 24, 2015 #4


    Staff: Mentor

    I think you misunderstood what phyzguy was saying. Here's what you wrote.
    Your substitution in the 3rd line above is wrong. sec(θ) = ##\sqrt{x^2-4x+5}##, not the ##x^2-4x+5## that you have. The 1/2 that you should be seeing comes from this square root.

    Also, I would have done a substitution with u = x - 2 before doing the trig substitution. Doing that makes life simpler, IMO.
  6. Mar 24, 2015 #5
    Whoops, yea I completely misunderstood and I will definitely try doing these with a u substitution for the x-a portion. Thanks!
  7. Mar 25, 2015 #6


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    Another approach is to split the integral up:

    ##\frac{x-1}{x^2-4x+5} = \frac{1}{2}\cdot \frac{2x - 4}{x^2-4x+5} + \frac{1}{x^2-4x+5}##

    Then ##\frac{1}{2}ln(x^2-4x+5)## comes easily out of the first term as you have an exact derivative in the numerator.
  8. Mar 25, 2015 #7
    Yea this is how wolfram alpha solved it but I figured it would be a good idea to understand both methods incase it's useful later on
  9. Mar 25, 2015 #8


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    Equivalently, completing the square, [tex]x^2- 4x+ 5= x^2- 4x+ 4+ 1= (x- 2)^2+ 1[/tex]. Let y= x- 2 so that dy= dx and y+ 1= x- 1.
    So [tex]\int \frac{x- 1}{x^2- 4x+ 5} dx= \int \frac{y+ 1}{y^2+ 1} dy= \int \frac{y}{y^2+ 1} dy+ \int\frac{1}{y^2+ 1} dy[/tex]
  10. Mar 26, 2015 #9
    Is it unusual that this integral can be solved in so many different ways?
  11. Mar 26, 2015 #10


    Staff: Mentor

    No, it happens fairly often. It's always true, though, that when you have two or more ways to evaluate an indefinite integral, the answers will differ by at most a constant. An example is ##\int sin(x)cos(x)dx##. One person can use a substitution u = sin(x), du = cos(x)dx. Another person can use a different substitution, u = cos(x), du = -sin(x)dx. When they integrate, they get what appear to be different answers, namely (1/2) sin2(x) + C and (-1/2)cos2(x) + C. However, since sin2(x) = 1 - cos2(x), it turns out that the two answers are different only by a constant.
  12. Mar 26, 2015 #11
    Okay thanks that makes sense.
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