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ggrree

  • 4 years ago

integral of dx/[1+sqrt x] (rewritten in reply)

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  1. ggrree
    • 4 years ago
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    \[\int\limits_{} {dx \over 1+ \sqrt x}\]

  2. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}\frac{1}{1+\sqrt{x}} dx\] \[\text{ let } \sqrt{x}=\tan^2(\theta) \] \[ x=\tan^4(\theta)\] => \[dx=4 \tan^3(\theta) \sec^2(\theta) d \theta\] So we have \[\int\limits_{}^{}\frac{4 \tan^3(\theta) \sec^2(\theta)}{1+\tan^2(\theta)} d \theta\] \[4 \int\limits_{}^{}\tan^3(\theta) d \theta\] That should help...

  3. myininaya
    • 4 years ago
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    We could try another way if you don't like this one...

  4. Mimi_x3
    • 4 years ago
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    \[\int\limits\frac{1}{1+\sqrt{x}} dx\] \[x = u^2 =>u=\sqrt{x} \] \[\frac{dx}{du} =2u =>dx=2u*du\] \[=>\int\limits\frac{1}{1+u} *2u*du =\int\limits\frac{u}{u+1} \] Might be easier..

  5. myininaya
    • 4 years ago
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    yep that's the one

  6. Mimi_x3
    • 4 years ago
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    woops, sorry

  7. Mimi_x3
    • 4 years ago
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    woops, i made a typo, the last one should be.. \[2\int\limits\frac{u}{u+1} du\]

  8. Chlorophyll
    • 4 years ago
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    Let u = sqrtx -> u^2 = x => 2udu = dx -> 2 Int ( udu/ u + 1) -> 2 Int [ 1 - 1/(u + 1 )] du = 2 [ u - ln(u + 1) ] = 2 ( sqrtx - ln(sqrtx + 1 ) + C

  9. ggrree
    • 4 years ago
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    thanks guys!

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