anonymous
  • anonymous
\[\int\limits {1 \over 3 \sqrt x +4x}dx--->x=u^2\]\[\int\limits {1 \over 3u + 4u^2}du\]How to carry on?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Can you show the steps?
anonymous
  • anonymous
u have missed 2udu iin the numerator and then u from denom and num cancels out and then u can assume 3+4u=z (or anything u wish) and then proceed to get the result the answer is i think (1/2)ln(3+4√x) +c
anonymous
  • anonymous
I don't know how to integrate this... can you help and show step by step?

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anonymous
  • anonymous
|dw:1333622702038:dw|
anonymous
  • anonymous
Why is 2udu=dx?
anonymous
  • anonymous
derivative of u^2 is 2u*u'
anonymous
  • anonymous
So, why does 2 go on the outside?
anonymous
  • anonymous
instead of 2u?
anonymous
  • anonymous
you can leave in the integral but it's a constant... you can pull it out of the integral... |dw:1333623126588:dw|
anonymous
  • anonymous
Ok. How do I carry on from your last step?
anonymous
  • anonymous
I can't see the right side..?
anonymous
  • anonymous
ok... wait... how about the previous drawings? could you see those?
anonymous
  • anonymous
not the one on the bottom right.
anonymous
  • anonymous
|dw:1333623470452:dw|
anonymous
  • anonymous
now you have to stick this back into the original problem...|dw:1333623691431:dw|
anonymous
  • anonymous
awww can you see that?
anonymous
  • anonymous
Not the full thing in black
anonymous
  • anonymous
I mean in the circle
anonymous
  • anonymous
the thing in the black circle is the problem we just worked out. so your answer is just that multiplied by 2. so you should have 1/2*ln |3+4sqrtx| + C
anonymous
  • anonymous
Thank you!

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