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anonymous

  • 5 years ago

Integrate (x^2 - 2x -1) / ( (x-1)^2 * (x^2 +1) ).

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  1. TuringTest
    • 5 years ago
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    factor the top and see what happens

  2. anonymous
    • 5 years ago
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    ok, let me try that...

  3. anonymous
    • 5 years ago
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    I dont think the top can be reduced any further

  4. TuringTest
    • 5 years ago
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    oh I read a plus... my mistake

  5. anonymous
    • 5 years ago
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    Partial fraction not helping?

  6. TuringTest
    • 5 years ago
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    yeah, I was trying to avoid it I completed the square on the top which let me split the integral into 1/(x^2+1)-2/[(x-1)^2(x^2+1)] I suppose you could do partial fractions on the second part, but it will be annoying

  7. anonymous
    • 5 years ago
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    I'm not sure how to begin splitting it for partial fractions. I'm wondering if I should write it as A/(x-1)^2 + B/(x^2 + 1) or A/(x-1) + B/(x-1) + C/(x^2 + 1).

  8. anonymous
    • 5 years ago
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    ok

  9. TuringTest
    • 5 years ago
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    I am trying A/(x-1) + B/(x-1)^2 + (Cx+D)/(x^2 + 1) you need the x on top for the squared x in the denom

  10. TuringTest
    • 5 years ago
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    do you see how I split the integral up in the first place by completing the square on top?

  11. anonymous
    • 5 years ago
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    yes

  12. TuringTest
    • 5 years ago
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    so then it's just this ugly PF thing... I'm gonna collect terms and give it to wolf to solve I think

  13. anonymous
    • 5 years ago
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    ok

  14. anonymous
    • 5 years ago
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    I'll be right back in 6 minutes.

  15. TuringTest
    • 5 years ago
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    ok I got A=1 B=-1 C=-1 D=0 so we should be able to do this now

  16. anonymous
    • 5 years ago
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    ok

  17. TuringTest
    • 5 years ago
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    yes, that works, I checked it with wolf. Want me to write out the whole thing or are you good?

  18. anonymous
    • 5 years ago
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    If you could write it out, i'd really appreciate it.

  19. TuringTest
    • 5 years ago
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    ok, but I'm not going to show all the partial fractions stuff... here we go:

  20. anonymous
    • 5 years ago
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    i'm still getting use to PF, so if u can, that would be great.

  21. anonymous
    • 5 years ago
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    but i understand the rest of what you wrote.

  22. TuringTest
    • 5 years ago
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    there are a few ways to do it, but here's the way I did it...

  23. TuringTest
    • 5 years ago
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    oh that was great everything I did just got erased. let's try that again...

  24. TuringTest
    • 5 years ago
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    \[\int\frac{x^2-2x-1}{(x-1)^2(x^2+1)}dx=\int\frac{(x-1)-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}dx\]just looking at the PF part:\[\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}=\frac{-2}{(x-1)^2(x^2+1)}\]multiply out by the denominator on the RHS to get rid of the fractions we obtain\[A(x-1)(x^2+1)+B(x-1)^2+(Cx+D)(x-1)^2=-2\]distribute the parentheses:\[A(x^3-x^2+x-1)+B(x^2+1)+C(x^3-2x^2+x)+D(x^2-2x+1)\]collect all like terms:\[x^3(A+C)+x^2(-A+B-2C+D)+x(A+C-2D)+(-A+B+D)\]now because this all equals -2 we know that the all the terms with x must have a coefficient of 0, so this leads us to the system\[x^3:A+C=0\]\[x^2:-A+B-2C+D=0\]\[x:A+C-2D=0\]\[-A+B+D=-2\]we can solve this in many ways, but here I will just give it to wolfram: http://www.wolframalpha.com/input/?i=A%2BC%3D0%2C-A%2BB-2C%2BD%3D0%2CA%2BC-2D%3D0%2C-A%2BB%2BD%3D-2 so now we know that\[A=1,B=-1,C=-1,D=0\]which leads to\[=\int\frac{1}{x^2+1}+\frac{1}{x-1}-\frac{1}{(x-1)^2}-\frac{x}{x^2+1}dx\]\[=\tan^{-1}x+\ln|x-1|+\frac{1}{x-1}+\frac{1}{2}\ln|x^2+1|+C\]

  25. TuringTest
    • 5 years ago
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    whew! that took a while. I bet there is a faster way to solve the system. For instance letting x=1 in the formula on the 5th line leads directly to\[2B=-2\to B=-1\]from which we can probably solve the system through substitution, but that was giving me a headache so I gave up. Better to use techniques from linear algebra like Gaussian elimination, or Cramer's rule, but you may not know those techniques yet.

  26. anonymous
    • 5 years ago
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    ok :). Thank you so much!

  27. TuringTest
    • 5 years ago
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    welcome :)

  28. TuringTest
    • 5 years ago
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    \[\int\frac{x^2-2x-1}{(x-1)^2(x^2+1)}dx=\int\frac{(x-1)-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}dx\]just looking at the PF part:\[\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}=\frac{-2}{(x-1)^2(x^2+1)}\]multiply out by the denominator on the RHS to get rid of the fractions we obtain\[A(x-1)(x^2+1)+B(x^2+1)+(Cx+D)(x-1)^2=-2\]distribute the parentheses:\[A(x^3-x^2+x-1)+B(x^2+1)+C(x^3-2x^2+x)+D(x^2-2x+1)\]collect all like terms:\[x^3(A+C)+x^2(-A+B-2C+D)+x(A+C-2D)+(-A+B+D)\]now because this all equals -2 we know that the all the terms with x must have a coefficient of 0, so this leads us to the system\[x^3:A+C=0\]\[x^2:-A+B-2C+D=0\]\[x:A+C-2D=0\]\[-A+B+D=-2\]we can solve this in many ways, but here I will just give it to wolfram: http://www.wolframalpha.com/input/?i=A%2BC%3D0%2C-A%2BB-2C%2BD%3D0%2CA%2BC-2D%3D0%2C-A%2BB%2BD%3D-2 so now we know that\[A=1,B=-1,C=-1,D=0\]which leads to\[=\int\frac{1}{x^2+1}+\frac{1}{x-1}-\frac{1}{(x-1)^2}-\frac{x}{x^2+1}dx\]\[=\tan^{-1}x+\ln|x-1|+\frac{1}{x-1}+\frac{1}{2}\ln|x^2+1|+C\] (there was a small typo above the last time, so I fixed it so as not to confuse you.)

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