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purplemouse

  • 4 years ago

Integral sqrt(x-2)/sqrt(x-1)

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  1. myininaya
    • 4 years ago
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    try u=sqrt(u-1) i will be right back

  2. myininaya
    • 4 years ago
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    \[u^2=x-1 => 2u u'=1 => u'=\frac{1}{2 u} => \frac{du}{dx}=\frac{1}{2 \sqrt{x-1}}\] \[2 du=\frac{dx}{\sqrt{x-1}}\] \[\int\limits_{}^{} \sqrt{u^2-1} \cdot 2 du=2 \int\limits_{}^{}\sqrt{u^2-1} du\] since sqrt(x-2)=sqrt(x-1-1)=sqrt(u^2-1) this part looks obvious to be to use a trig sub now

  3. purplemouse
    • 4 years ago
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    Thanks a bunch!

  4. myininaya
    • 4 years ago
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    :)

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