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RolyPoly

  • 2 years ago

This is the first dish \[\int \frac{x^2}{1-x^2}dx\] Apart from resolving that into partial fractions, what is a better way to do it?

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  1. RolyPoly
    • 2 years ago
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    let x=sinu \[\int \frac{x^2}{1-x^2}dx = \int\frac{sin^2ucosu}{cos^2u}du=\int(sec^2u-cosu)du=tanu-sinu+C\]?

  2. RolyPoly
    • 2 years ago
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    Last line there: tanu -sinu +C \[=\frac{x}{x^2+1}-x+C\] What am I doing here :S

  3. RolyPoly
    • 2 years ago
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    \[\frac{sin^2ucosu}{cos^2u}=\frac{(1-cos^2u)cosu}{cos^2u} = sec^2u-cosu\]

  4. hartnn
    • 2 years ago
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    tan u is x/ sqrt(1-x^2)

  5. RolyPoly
    • 2 years ago
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    Oh.. My mistake! \[=\frac{x}{1-x^2} -x+C\]

  6. RolyPoly
    • 2 years ago
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    Alright, let's go to the main dish :S

  7. hartnn
    • 2 years ago
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    and its not sec^2 , check again

  8. hartnn
    • 2 years ago
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    sec-cos

  9. RolyPoly
    • 2 years ago
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    secu! I'd better do it all over again!!! \[\int \frac{x^2}{1-x^2}dx \]\[= \int\frac{sin^2ucosu}{cos^2u}du\]\[=\int(secu-cosu)du\]\[=\ln|secu+tanu|-sinu+C\]

  10. hartnn
    • 2 years ago
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    not a better way......

  11. RolyPoly
    • 2 years ago
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    I know, I just try whatever I can.. Is there a better way then?

  12. hartnn
    • 2 years ago
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    i don't think,partial is best....

  13. RolyPoly
    • 2 years ago
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    \[\int \frac{x^2}{1-x^2}dx\]\[ = \int (\frac{1}{2(x+1)}-\frac{1}{2(x-1)}-1)dx\]\[=\frac{1}{2} ln|x+1| -\frac{1}{2}ln|x-1|-x+C\]\[=\frac{1}{2}ln|\frac{x+1}{x-1}|-x+C\]

  14. hartnn
    • 2 years ago
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    yup, thats correct.....

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