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anonymous

  • one year ago

make a substitution to express the integrand as a rational function and then evaluate the integral how do i do this? problem inside!! thanks!!

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  1. anonymous
    • one year ago
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    \[\int\limits_{} \frac{ dx }{ 2\sqrt{x+3}+x }\]

  2. Astrophysics
    • one year ago
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    What have you tried?

  3. anonymous
    • one year ago
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    i am unsure of which concept i am using and how to apply am i doing the u= ____ du=____ ?

  4. Astrophysics
    • one year ago
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    Yes, when they ask for a substitution you can make u = something

  5. anonymous
    • one year ago
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    okay, would it be u=2 (sqrt x+3) +x and du = dx ?

  6. anonymous
    • one year ago
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    i am very confused :(

  7. Astrophysics
    • one year ago
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    Yes that's a good substitution! But when you have to take the derivative with respect to x that's why you have \[\frac{ du }{ dx }\]

  8. anonymous
    • one year ago
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    ohh okay, so what would i be using as substitution in this case then?

  9. anonymous
    • one year ago
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    can i use this? u=2 (sqrt x+3) +x and du = dx ?

  10. Astrophysics
    • one year ago
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    You will be using \[u=\sqrt{x+3}\] what's the derivative of this?

  11. anonymous
    • one year ago
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    would it be 1 ----- 2 sqrt(x+3) ?

  12. anonymous
    • one year ago
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    and if so, that would be du?

  13. Astrophysics
    • one year ago
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    Yup, but note we have that +x in the denominator so you have solve for x here as well \[u=\sqrt{x+3}\] solve for x

  14. anonymous
    • one year ago
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    \[\sqrt{x+3}=u,x+3=u^2,x=u^2-3,dx=2udu\]

  15. Astrophysics
    • one year ago
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    \[du = \frac{ 1 }{ 2\sqrt{x+3} }dx\]

  16. anonymous
    • one year ago
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    ooh okie!! so what happens next? :/

  17. anonymous
    • one year ago
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    do i still need to find more substitutions?

  18. Astrophysics
    • one year ago
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    Well I asked you to solve for x, but surji already did it for you... now just plug it all in.

  19. anonymous
    • one year ago
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    ohh okay, so i plug into the original equation?

  20. Astrophysics
    • one year ago
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    Yes, that's why it's called substitution :)

  21. anonymous
    • one year ago
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    like this? \[\int\limits_{} \frac{ 2udu }{ 2\sqrt{(u^2-3)+3}+(u^2-3) }\]

  22. Astrophysics
    • one year ago
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    Nope, lets go over this slowly

  23. anonymous
    • one year ago
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    aww:( okay!!

  24. Astrophysics
    • one year ago
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    So our original integral is \[\int\limits \frac{ dx }{ 2\sqrt{x+3}+x }\] we made the substitution \[u= \sqrt{x+3} \implies du = \frac{ 1 }{ 2\sqrt{x+3} }dx\] but notice the integrand still has that +x in the denominator, and we can't have that when we want everything respect to u, so we look at our original substitution and see if there is a way we can get rid of the x, \[u=\sqrt{x+3} \implies x = u^2-3\] so we plug that in where that +x is, but note we can simplify our derivative \[du = \frac{ 1 }{ 2\sqrt{x+3} }dx \implies du = \frac{ 1 }{ 2u }dx \implies 2udu = dx\]

  25. Astrophysics
    • one year ago
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    Now try and plug it all in

  26. anonymous
    • one year ago
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    whoahh okay, let me try :)

  27. anonymous
    • one year ago
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    \[\int\limits_{} \frac{ 2udu }{ 2u+(u^2-3) }\]

  28. anonymous
    • one year ago
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    this?

  29. Astrophysics
    • one year ago
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    Yes, exactly!

  30. anonymous
    • one year ago
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    ooh okay yay!! so what do i do now?

  31. Astrophysics
    • one year ago
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    Now you can evaluate it

  32. anonymous
    • one year ago
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    how can i do that? :/ sorry, i am having a hard time understanding calc :( i follow so far right now, but i do not understand what I am supposed to do when you say evaluate? is it finding the antiderivative?

  33. Astrophysics
    • one year ago
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    Yes, when you integrate you find the antiderivative. You will have to use partial fractions

  34. anonymous
    • one year ago
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    okay, i am not very good with partial fractions :( could you possibly guide me through the steps that I would do here? I just don't know what i do here :/ find A,B and C?

  35. Astrophysics
    • one year ago
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    Well partial fractions can only be learnt through practice, you have to practice in calculus if you want to pass, so you should practice these problems after. Lets start of by looking at the integrand \[2 \int\limits \frac{ u }{ u^2+2u-3 }du\] lets factor the denominator what do you get?

  36. anonymous
    • one year ago
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    i get (u+1)-1 ? :/

  37. Astrophysics
    • one year ago
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    Factor the denominator u^2+2u-3

  38. anonymous
    • one year ago
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    okay, so (u+1)(u-4) ? :/

  39. anonymous
    • one year ago
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    (u+1)^2 + 1 ?

  40. anonymous
    • one year ago
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    i don't really know :(

  41. Astrophysics
    • one year ago
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    Ok so you will have to review factoring as well, you do this in high school, you should have got (u+3)(u-1)

  42. Astrophysics
    • one year ago
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    So we have now \[2 \int\limits \frac{ u }{ (u+3)(u-1) }du\] this is our integral now

  43. Astrophysics
    • one year ago
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    So for partial fractions we set it up as such \[\frac{ u }{ (u+3)(u-1) } = \frac{ A}{ (u+3) }+\frac{ B }{ (u-1) }\]

  44. anonymous
    • one year ago
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    ohhh okay, oops, sorry thought i had to do the thing where you add a 1 to the end or something :/

  45. anonymous
    • one year ago
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    okay! and how do we solve for A and B ?

  46. Astrophysics
    • one year ago
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    No you're thinking of completing the square, which is also helpful for some problems

  47. anonymous
    • one year ago
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    ohh okay, oopsies :P

  48. Astrophysics
    • one year ago
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    Multiply both sides by (u+3)(u-1) what do you get

  49. anonymous
    • one year ago
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    A(u-1) and B(u+3) ?

  50. Astrophysics
    • one year ago
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    Yes, but you must write it as \[u=A(u-1)+B(u+3)\]

  51. Astrophysics
    • one year ago
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    Now distribute the A and B

  52. anonymous
    • one year ago
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    ohh okay!! distribute like Au-A + Bu+3B ?

  53. Astrophysics
    • one year ago
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    Yeah, now we look at what we have on the left side, \[u=(A+B)+(-A+3B)\] now we can set up a system of equations \[A+B=1\] \[-A+3B=0\] I think you should be able to do the rest

  54. anonymous
    • one year ago
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    okay!! err, still a bit confused :/ what would i be doing next? do i sub in to the A/u+3 + B/u-1 ?

  55. anonymous
    • one year ago
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    with A=1 and B=0? :/

  56. Astrophysics
    • one year ago
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    This is just a regular system of equations solve for A and B

  57. Astrophysics
    • one year ago
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    You can use any method, I recommend just adding as it's the most obvious in this case

  58. anonymous
    • one year ago
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    ok, so we get A=3B --> 3B+B =1 =4B = 1 -->B=1/4 then A=3/4 ?

  59. anonymous
    • one year ago
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    did i do that correctly?

  60. Astrophysics
    • one year ago
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    Yeah, looks good!

  61. anonymous
    • one year ago
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    yay! what do i do now?

  62. Astrophysics
    • one year ago
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    \[\frac{ u }{ (u+3)(u-1) } = \frac{ 3 }{ 4(u+3) }+\frac{ 1 }{ 4(u-1) }\] now this will become your integral so we have \[2 \int\limits \frac{ 3 }{ 4(u+3) }+\frac{ 1 }{ 4(u-1) }du\] this should be easy now, just evaluate

  63. anonymous
    • one year ago
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    so would i get this?\[\frac{ 1 }{ 2 } \log(1-u) + 3\log(u+3) + C \]

  64. Astrophysics
    • one year ago
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    \[2(3/4\ln|u+3|+14\ln|u-1|)+C\] but you have to sub back the u substitution we made

  65. Astrophysics
    • one year ago
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    1/4 not 14

  66. anonymous
    • one year ago
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    okay! so we get this? \[2(3/4\ln \left| 5/4+3 \right|+ 1/4\ln \left| 5/4-1 \right| +C\]

  67. Astrophysics
    • one year ago
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    5/4?

  68. Astrophysics
    • one year ago
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    Curious, where did you get the 5/4 from?

  69. anonymous
    • one year ago
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    i am not quite sure if i calculated correctly? u=A+B -A +3B ? =(3/4)+(1/4)-(3/4)+3(1/4) =1 ? oops :P

  70. Astrophysics
    • one year ago
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    Noooo, all we have to do is go back and see \[u= \sqrt{x+3}\] \[\implies 2\left( \frac{ 3 }{ 4 }\ln|\sqrt{x+3}+3|+\frac{ 1 }{ 4 }\ln|\sqrt{x+3}-1| \right)+C\]

  71. anonymous
    • one year ago
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    ohhh i see so what do we do next?

  72. Astrophysics
    • one year ago
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    We're done, the problem not long enough? :P

  73. anonymous
    • one year ago
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    hahaha ohhh okay :P yay!! thank you so ugh!! it was so long!! but i think i understand more of it now :) thanks so much!!! I'm having trouble understanding the concepts for calc right now so i appreciate your time and help!!

  74. Astrophysics
    • one year ago
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    Np, integrals require lots of practice, I haven't mastered them myself xD, they can be tricky, they're not as easy as derivatives! So keep practicing!

  75. anonymous
    • one year ago
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    :)

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