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anonymous

  • 4 years ago

find the integral using long division...?

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{}^{}(27x^2)\div (3x+4)^3 dx\]

  2. anonymous
    • 4 years ago
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    oh ok

  3. myininaya
    • 4 years ago
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    Do we have to do long division? Can we use partial fractions instead?

  4. myininaya
    • 4 years ago
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    It actually says do long division?

  5. anonymous
    • 4 years ago
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    Long division is going to get you nowhere in this case. Notice the power of the denominator is bigger than that of the numerator.

  6. myininaya
    • 4 years ago
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    Oh that is true

  7. myininaya
    • 4 years ago
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    good eye

  8. anonymous
    • 4 years ago
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    ok, so how would you do it with partial fractions?

  9. anonymous
    • 4 years ago
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    the other parts of the question i had to use long division so i figured it would be the same throughout.. guess thats why i got confused.

  10. anonymous
    • 4 years ago
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    Would you like a step by step?

  11. anonymous
    • 4 years ago
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    if you could, that would help a lot!

  12. anonymous
    • 4 years ago
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    No problem at all, first: how familiar are you with partial fractions?

  13. anonymous
    • 4 years ago
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    i'm not familiar with the name but once i see it i will probably remember..

  14. anonymous
    • 4 years ago
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    i don't think we have done partial fractions, i just looked it up on wolfram alpha and it doesn't look familiar

  15. anonymous
    • 4 years ago
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    is there any other way to solve it?

  16. anonymous
    • 4 years ago
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    Okay, well in general, partial fractions is a way to take rational expression: \[\frac{P(x)}{Q(x)}\] And write it as a sum of lower order terms. Think of it as "undoing" a common denominator. So we have: \[\frac{27x^2}{(3x+4)^3}=\frac{A}{3x+4}+\frac{B}{(3x+4)^2}+\frac{C}{(3x+4)^3} \implies \] \[A(3x+4)^2+B(3x+4)+C=27x^2\]

  17. anonymous
    • 4 years ago
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    ok now it looks familiar!!

  18. anonymous
    • 4 years ago
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    so how would i get A B and C?

  19. anonymous
    • 4 years ago
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    Okay, to do that we simply multiply it all out, and set the coefficients of the terms equal. Sounds odd but let me show you what I mean. First, you notice that if you let x=-4/3 that the 3x+4 terms go to zero? So lets do that, let x=-4/3 then we get: \[A(0)^2+B(0)+C=27(-4/3)^2 \implies C=27(16/9)=3*16=48\] Now we know C! So we have: \[A(3x+4)^2+(3x+4)B+48=27x^2\] Now we can do my method, so we have: \[A(9x^2+16+24x)+B(3x+4)+48=27x^2 \] Now we can deduce what we want to know. Looking at this, you realize for the sides to be equal, the coefficient 27 must equal the coefficients in front of the x^2 terms on the left. So for this example: \[27=9A \implies A=3; 24A+3B=0, A=3 \implies B=-24\] Thus we have: \[\int\limits \frac{27x^2}{(3x+4)^3}dx=\int\limits \left[ \frac{3}{3x+4}-\frac{24}{(3x+4)^2}+\frac{48}{(3x+4)^3} \right]dx\] Follow me to here?

  20. anonymous
    • 4 years ago
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    Is it ok if i wrote it:\[27\int\limits_{}^{}1\div(9(3x+4)-8\div9(3x+4)^2+16\div9(3x+4)^3 dx\]

  21. anonymous
    • 4 years ago
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    i took out constant first are my numbers correct?

  22. anonymous
    • 4 years ago
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    Yeah, it would be just (constant)/27. So I believe so. Do you know how to integrate each of these terms?

  23. anonymous
    • 4 years ago
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    well what would i use for my u value if i were to use u-substitution?

  24. anonymous
    • 4 years ago
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    so that i only have to substitute once.

  25. anonymous
    • 4 years ago
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    For all of them let u=3x+4; (1/3)du=dx

  26. anonymous
    • 4 years ago
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    but i will still have to substitute again after

  27. anonymous
    • 4 years ago
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    Not quite: \[9 \int\limits \frac{du}{9u}-8 \int\limits \frac{du}{27u^2}+16 \int\limits \frac{du}{27u^3}\] See if you agree. Note: I used your numbers so the constants may be slightly off of what I have above^^

  28. anonymous
    • 4 years ago
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    Thats doing all of it in 1 substitution. Obviously when you integrate it you'll have a function of u, then you replug in what your u-sub was.

  29. anonymous
    • 4 years ago
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    i got \[(u^2\div2)+(8\div27u)-(8\div27u^2)+ c] is that correct?

  30. anonymous
    • 4 years ago
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    \[(u^2/2)+(8/27u)-(8/27u^2)+c\]**

  31. anonymous
    • 4 years ago
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    then plug in u-value?

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