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agentx5

  • 3 years ago

I need some help with checking this answer please: \[\large\int\limits_4^5 \frac{x^3-3 x^2-9}{x^3-3 x^2} dx = x-\frac{3}{x}-\ln|3-x|+\ln|x|+C\] I got this far using the partial fractions technique. I went to check my answer that I got on paper but Wolfram gives some craziness with hyperbolic tangents that I don't understand. So what should the answer be and wouldn't the natural log of 3-x be an issue because it gives non-real values? The correct answer MUST have real values and be in terms of constants and logs. If somebody could show me what's going on here it would be great :-)

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  1. agentx5
    • 3 years ago
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    For my partial fractions I got: \[x^3-3x^2-9=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-3}\] The A part turns into ln|x|, the B part turns into 3/x, the C part turns into -ln|3-x|

  2. helder_edwin
    • 3 years ago
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    \[ \Large \int\frac{x^3-3x^2-9}{x^3-3x^2}\,dx= \int\,dx-\int\frac{9}{x^2(x-3)}\,dx \] so you have to solve \[ \Large \frac{9}{x^2(x-3)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-3} \]

  3. Valpey
    • 3 years ago
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    Well, for one thing you have written the natural log of the absolute value of 3-x.

  4. agentx5
    • 3 years ago
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    Yeah I used long division initially to pull out this: \[1 + \frac{-9}{x^3-3x^2}\]

  5. agentx5
    • 3 years ago
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    Integral of one with respect to x is just x, logically

  6. agentx5
    • 3 years ago
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    I think I mistyped my second post, sry @helder_edwin

  7. agentx5
    • 3 years ago
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    @Valpey oh yeah... so it's always a positive...

  8. agentx5
    • 3 years ago
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    ln|3-(4)| = ln|-1| = ln 1 = 0

  9. agentx5
    • 3 years ago
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    That's a nice simplification

  10. agentx5
    • 3 years ago
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    /me waves hello @asnaseer :-) ty for stopping by!

  11. asnaseer
    • 3 years ago
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    you didn't need to do long division to simplify this in the first place. you could just have done this:\[\frac{x^3-3x^2-9}{x^3-3x^2}=\frac{x^3-3x^2}{x^3-3x^2}-\frac{9}{x^3-3x^2}=1-\frac{9}{x^3-3x^2}\]

  12. asnaseer
    • 3 years ago
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    secondly, the logs should not contain the absolute value symbols around them

  13. agentx5
    • 3 years ago
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    Good point @asnaseer and why not? Without it wouldn't I have a non-real answer issue when I go to evaluate?

  14. asnaseer
    • 3 years ago
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    you should have done this:\[\int\limits_4^5 \frac{x^3-3 x^2-9}{x^3-3 x^2} dx =[ x-\frac{3}{x}-\ln(3-x)+\ln(x)]_4^5\]\[\qquad=[x-\frac{3}{x}+ln(\frac{x}{3-x})]_4^5\]\[\qquad=(5-\frac{3}{5}+ln(-\frac{5}{2}))-(4-\frac{3}{4}+ln(-\frac{4}{1}))\]\[\qquad=5-\frac{3}{5}-4+\frac{3}{4}+ln(-\frac{5}{2}\div-\frac{4}{1})\]

  15. asnaseer
    • 3 years ago
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    and the minuses cancel out inside the logs

  16. agentx5
    • 3 years ago
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    *reading carefully*

  17. asnaseer
    • 3 years ago
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    BTW: waves hello back to @agentx5 :)

  18. agentx5
    • 3 years ago
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    Oh... properties of logs... \[\ln|a|-\ln|b|=\ln|\frac{a}{b}|\]

  19. asnaseer
    • 3 years ago
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    yes - but you shouldn't use the '|' symbols here

  20. asnaseer
    • 3 years ago
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    as they usually indicate absolute value

  21. agentx5
    • 3 years ago
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    \[\qquad=5-\frac{3}{5}-4+\frac{3}{4}+ln(-\frac{5}{2}\div-\frac{4}{1}) \approx 3.45258\] Disagrees with this check: http://www.wolframalpha.com/input/?i=integrate+from+4+to+5+for+%28x^3-3x^2-9%29%2F%28x^3-3x^2%29 \(\approx \frac{17}{25}\)

  22. asnaseer
    • 3 years ago
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    I work it out to be exactly the same as the answer wolfram gives

  23. asnaseer
    • 3 years ago
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    I get:\[\frac{23}{20}+ln(\frac{5}{8})\approx 0.68\]

  24. agentx5
    • 3 years ago
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    Ugh I see now I used the wrong base log on my calculator... /facepalm

  25. agentx5
    • 3 years ago
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    Derp.

  26. asnaseer
    • 3 years ago
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    np :)

  27. agentx5
    • 3 years ago
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    Thank you so much you were a great help! @TuringTest give that claymation-man a medal ^_^

  28. asnaseer
    • 3 years ago
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    you are more than welcome my friend! :)

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