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anonymous

  • 5 years ago

what is the integration of 1/{(x+2)(sqr of x+3)} ?

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  1. anonymous
    • 5 years ago
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    Is that suppose to be 1/{(x+2)(x+3)(x+3)} What is squared?

  2. anonymous
    • 5 years ago
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    no...it is \[1/({(x+2)\sqrt{x+3}})\]

  3. anonymous
    • 5 years ago
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    how can this be solved?

  4. anonymous
    • 5 years ago
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    What Calculus class are you in?

  5. anonymous
    • 5 years ago
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    this can be solved using u subs?

  6. anonymous
    • 5 years ago
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    What Calculus class are you in?

  7. anonymous
    • 5 years ago
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    what?

  8. anonymous
    • 5 years ago
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    Are you doing this for fun or a class?

  9. anonymous
    • 5 years ago
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    class of course...wanted to learn

  10. anonymous
    • 5 years ago
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    What type of math class? Clause it depends on what way.

  11. anonymous
    • 5 years ago
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    i had learnt basic calculus...it is a bit advanced this time

  12. anonymous
    • 5 years ago
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    "learned" so you're in basic calculus, trying to get ahead?

  13. anonymous
    • 5 years ago
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    yeah

  14. anonymous
    • 5 years ago
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    OK, what ways do you know how to use?

  15. anonymous
    • 5 years ago
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    for this prob, i guess we can use integration by parts..right? but i always get confused in which term to take as 'u' and which as 'v'....i need some hints..

  16. anonymous
    • 5 years ago
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    In order to do that your going to need to change some parts and move them around a little. I would change the square root to a power and take it out from the demomatator.

  17. anonymous
    • 5 years ago
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    like denominator=(x+2).(x+3)\[^{1/2}\]

  18. anonymous
    • 5 years ago
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    (x+2).(x+3)1/2

  19. anonymous
    • 5 years ago
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    with the 1/2 on the (x+3) yes then pull it out so it becomes a product rule looking equation

  20. anonymous
    • 5 years ago
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    then it would look like \[1\div ((x+2).(x+3)^{1/2})\]

  21. anonymous
    • 5 years ago
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    \[1\div \left( x+2 \right)\left( x+3 \right)^{1/2}\] to \[\left( x+3 \right)^{-1/2} \] next to 1÷(x+2)

  22. anonymous
    • 5 years ago
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    okay so now it is \[\int\limits_{}^{} [(x+3)^{-1/2}]/(x+2)\]

  23. anonymous
    • 5 years ago
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    you could look at it that way but it might be more helpful if you had that in two separate parts. might be easier to work with

  24. anonymous
    • 5 years ago
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    Think product rule versus quot. rule

  25. anonymous
    • 5 years ago
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    so can you solve this now?..i am solving it right now...plz do solve and let me know the answer..

  26. anonymous
    • 5 years ago
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    This has taken longer than I planned, and I'm about to be sleeping on my keyboard. I hope this helps

  27. anonymous
    • 5 years ago
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    correct me if i am wrong , but after intr. for the first time i get, 2\[\sqrt{x+3}/(x+2)-(2 \int\limits_{}^{}\sqrt{x+3} . \log(x+3))dx\]

  28. anonymous
    • 5 years ago
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    I think your missing a /2 in the first part, but my brain is turning off. night, sorry but I'm just too tired to think much more, worst case I'll check back tomorrow.

  29. anonymous
    • 5 years ago
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    okay....i'll try by then....

  30. anonymous
    • 5 years ago
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    did u get the solution?

  31. anonymous
    • 5 years ago
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    yes, I'm going to type the whole solution out for you... hold on

  32. anonymous
    • 5 years ago
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    thanks

  33. anonymous
    • 5 years ago
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    \[\int\limits \frac{1}{(x+2)\sqrt {x+3}}dx \rightarrow u=\sqrt{x+3}, 2du=\frac{1}{\sqrt{x+3}}dx\] \[2 \int\limits \frac{1}{x+2}du \rightarrow now:u^2-1=x+2, so \rightarrow 2 \int\limits \frac{1}{u^2-1}du\] Use partial fraction decomposition: \[\frac{1}{u^2-1}=\frac {1}{(u+1)(u-1)}= \frac{A}{u+1}+\frac {B}{u-1}\] \[A(u-1)+B(u+1)=1\rightarrow Au-A+Bu+B=1\] \[(B+A)u=0u, B-A=1\rightarrow so: B+A=0, B-A=1\] solve for B and A: \[B=\frac{1}{2}, A=- \frac{1}{2}\] \[2 \int\limits\limits \frac{1}{2} (\frac{1}{u-1}-\frac{1}{u+1})du \rightarrow \int\limits \frac{1}{u-1}du- \int\limits \frac{1}{u+1}du\] \[\int\limits\limits \frac{1}{u-1}du \rightarrow v=u-1, dv=du\] \[\rightarrow \int\limits\limits \frac{1}{v}dv=\ln(v)=\ln(u-1)\] \[\int\limits \frac {1}{u+1}du \rightarrow z=u+1, dz=du\] \[\rightarrow \int\limits \frac{1}{z}dz=\ln(z)=\ln(u+1)\] so:\[\ln(u-1)-\ln(u+1)+C \rightarrow \ln \left| \frac {u-1}{u+1} \right|+C\] From the begining we know what u is, so: \[since:u=\sqrt{x+3}\rightarrow \ln \left| \frac{\sqrt{x+3}-1}{\sqrt{x+3}+1}\right|+C\]

  34. anonymous
    • 5 years ago
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    how did you take : u2−1=x+2 ?

  35. anonymous
    • 5 years ago
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    like this: \[u=\sqrt{x+3}\rightarrow u^2=(\sqrt{x+3})^2\rightarrow u^2=x+3\rightarrow u^2-1=x+2\]

  36. anonymous
    • 5 years ago
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    ohh yeah..thanks nadeem

  37. anonymous
    • 5 years ago
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    there's one more question...if u can answer...its regarding partial derivatives..can u?

  38. anonymous
    • 5 years ago
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    sure

  39. anonymous
    • 5 years ago
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    its just 2 places above this post.. which is like fxy =(-y sin xy)...so on...plz move to that post...

  40. anonymous
    • 5 years ago
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    i got it :)

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