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anonymous

  • 5 years ago

\[\int\limits_{}^{} (6x-11 / x+2) dx\]

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  1. amistre64
    • 5 years ago
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    nice job with the equation writing skill...kudos ;)

  2. amistre64
    • 5 years ago
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    frac{top}{bottom} makes a clean fraction

  3. anonymous
    • 5 years ago
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    You need to substitute x for x=u+2 and then solve (dx=du since d/dy of +2 is zero).

  4. anonymous
    • 5 years ago
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    Err... that's not d/dy of +2 but the derivative of +2

  5. amistre64
    • 5 years ago
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    \[\frac{top}{australia}\]

  6. amistre64
    • 5 years ago
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    \[\int \frac{6x-11}{x+2}dx\]

  7. anonymous
    • 5 years ago
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    thanks for the lesson in writing equations >:(

  8. anonymous
    • 5 years ago
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    Do you get what I wrote msunprecedented?

  9. anonymous
    • 5 years ago
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    no, i did not

  10. amistre64
    • 5 years ago
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    lol.... I was impressed by your talent ...still am :)

  11. amistre64
    • 5 years ago
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    do you want the integral of 3 seperate terms; or is what I posted a correct interpretation of it?

  12. anonymous
    • 5 years ago
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    no its correct \[\int\limits_{}^{} \frac {6x-11} {x+2} dx\]

  13. amistre64
    • 5 years ago
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    ... factoring might be possible but we would have to find out what value had been canceled before hand.... and then do partial decomposition as a possibility

  14. anonymous
    • 5 years ago
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    Actually, this is really simple if you view the fraction like this: \[6x/(x+2) - (11/x+2)\] Now that it's in this form, you split the integral into two : \[\int\limits_{?}^{?}6x/(x+2)dx - \int\limits_{?}^{?}11/(6x+2)dx\] This should be easier to integrate... let me know if you need more help

  15. amistre64
    • 5 years ago
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    the left is easy now :) the right still has a straggler.... that "x" up top can get bothersome

  16. amistre64
    • 5 years ago
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    and by left I mean right ;)

  17. amistre64
    • 5 years ago
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    would the uv substitution work well there?

  18. anonymous
    • 5 years ago
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    uv substitution wouldn't work because I don't think you could factor x+2. (I can't see a way to factor anyway)

  19. amistre64
    • 5 years ago
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    6x goes to zero pretty quick like; but I get ln(x+2) and get lost on integrateing that

  20. amistre64
    • 5 years ago
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  21. anonymous
    • 5 years ago
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    Hmm.. I think integration by substitution works here. If you let u=x+2, you can get the right equal to:\[\int\limits_{?}^{?} (6u-12)/u *du\] Then you split the fraction one more time

  22. anonymous
    • 5 years ago
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    I mean the left when I say the "right" above

  23. amistre64
    • 5 years ago
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    lol..... stage left ...STAGE left!! :)

  24. anonymous
    • 5 years ago
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    Yikes! I'm soo confused! U-sub here? Integration by parts there? :(

  25. anonymous
    • 5 years ago
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    msunprecedented, have you learnt how to do integration by substitution yet?

  26. anonymous
    • 5 years ago
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    I believe not.

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