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anonymous

  • 4 years ago

Which method can help me to do the integral of (x^2+1)/(1-x^2)?

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  1. AravindG
    • 4 years ago
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    quotient rule may be

  2. nenadmatematika
    • 4 years ago
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    write x^2+1 as x^2-1+2, make two basic integrals and solve

  3. anonymous
    • 4 years ago
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    Because I just learned in school to do a partial fraction decomposition but this somehow does not work out very well. I´ll use your tip nenadmatematika.

  4. dumbcow
    • 4 years ago
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    ha i was just going to suggest partial fractions 1-x^2 = (1-x)(1+x)

  5. dumbcow
    • 4 years ago
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    oh right split it into 2 fractions first

  6. nenadmatematika
    • 4 years ago
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    \[=\int\limits_{?}^{?}(x^2-1)/(1-x^2)dx+\int\limits_{?}^{?}2/(1-x^2)=-\int\limits_{?}^{?}dx+2\int\limits_{}^{?}dx/(1-x^2)\]\[=-x+\ln \left| (1+x)/(1-x) \right|+c\]

  7. anonymous
    • 4 years ago
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    This kind of idea comes from experience I guess?

  8. nenadmatematika
    • 4 years ago
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    yes, it takes some practice to get the routine, but consider it very easy integral...:D

  9. anonymous
    • 4 years ago
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    ok, thx for your help :)

  10. nenadmatematika
    • 4 years ago
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    if you've got some more problems with integrals hit me now :D

  11. TuringTest
    • 4 years ago
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    Do you want to see it with partial fractions? it's not very hard

  12. anonymous
    • 4 years ago
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    Yes, might be a nice to see a different approach.

  13. TuringTest
    • 4 years ago
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    well actually, I guess nenad did use them it do the second integral (or at least I don't see how he did without them) another way to get to that point it to rearrange the fraction so you can perform long division\[\frac{1+x^2}{1-x^2}=\frac{-x^2-1}{x^2-1}=-1-\frac2{x^2-1}\]now doing partial fractions on the second term\[\frac{-2}{x^2-1}=\frac A{x-1}+\frac B{x+1}\]\[A(x+1)+B(x-1)=-2\]\[A=-1,B=1\]so we wind up with\[\int-1+\frac1{x-1}-\frac1{x+1}dx\]again

  14. nenadmatematika
    • 4 years ago
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    well, that's nice but you would you make that second part complicated when you know that it's a basic integral: \[\int\limits_{?}^{?}dx/(1-x^2)=(1/2)\ln \left| (1+x)/(1-x)\right|\]

  15. TuringTest
    • 4 years ago
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    I didn't know that form, but if they are supposed to use partial fractions in the problem they must do it here.

  16. nenadmatematika
    • 4 years ago
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    OK, I didn't know that they must do it like that :D

  17. anonymous
    • 4 years ago
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    Can I do this partial fractions stuff also without changing x^2+1 to x^2-1+2?

  18. TuringTest
    • 4 years ago
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    the order of the numerator has to be smaller than that in the denominator to do partial fractions, so you need to make that happen somehow long division is a very standard procedure in these cases, but you can also do what nenad did frequently.

  19. TuringTest
    • 4 years ago
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    the degree of the numerator*

  20. anonymous
    • 4 years ago
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    Ah, Thank you. We only talked in school about the cases where the degree is higher or lower but never about what it means when they have the same degree. :( My teacher sucks... Thank you TuringTest for your help and explanation.

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