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anonymous

  • 4 years ago

find the anti-derivative

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{}^{?}2x/(x-1)^{2}\] the ? doesn't mean anything

  2. anonymous
    • 4 years ago
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    try \[u=x-1\] \[du=dx\] and \[x=u-1\] so \[2x = 2u-2\]

  3. anonymous
    • 4 years ago
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    oops sorry if \[u=x-1\] then \[x=u+1\] and \[2x=2u+2\] my mistake

  4. anonymous
    • 4 years ago
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    good from there?

  5. anonymous
    • 4 years ago
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    yes but after that the (2x+2) *u^-2 would equal 2u^-1 +2u^-2 and ln is required for the -1 exponent and that is what i am having trouble with

  6. anonymous
    • 4 years ago
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    you should have only "u" involved. as follows \[\int\frac{2u+2}{u^2}du\] then break it in to two parts

  7. anonymous
    • 4 years ago
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    \[\int \frac{2u}{u^2}du+\int \frac{2}{u^2}du\] first one becomes \[\int \frac{2}{u}du\] whose "anti derivative" is the log

  8. anonymous
    • 4 years ago
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    oooh i see what you are asking. you cannot use the power rule backwards when integrating \[\int \frac{1}{x}dx\] that is \[\frac{1}{x}dx=\ln(x)\]

  9. anonymous
    • 4 years ago
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    ohhhhhh ok i understand that now thanks

  10. anonymous
    • 4 years ago
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    i meant of course \[\int \frac{1}{x}dx=\ln(x)\]

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