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anonymous

  • 5 years ago

Indefinite Integral Help int (x/(2x+1) dx I keep getting (x/2)-((1/4)ln|2x+1|)+1/4+C Book says ((1/2)x)-((1/4)ln|2x+1|)+C. So when I get a constant as part of the answer when trying to find the indefinite integral, do I just consider it as part of the constant C? Is that why I'm off by +1/4?

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  1. TuringTest
    • 5 years ago
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    yes, 1/4 is absorbed into C, because C is unknown that will happen a lot in differential equations, so get used to it

  2. Akshay_Budhkar
    • 5 years ago
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    yes whenever you have another constant, you ought to add it into C. The C absorbs it.. If you are having marks for steps then this step will count for marks as well ( maybe 1/2 mark or so if u r scoring full elsewhere)

  3. TuringTest
    • 5 years ago
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    wolfram keeps the 1 though ironically http://www.wolframalpha.com/input/?i=int%20(x%2F(2x%2B1)%20dx&t=crmtb01 but they left it in the parentheses, so I guess they feel it's justified

  4. Akshay_Budhkar
    • 5 years ago
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    but in the alternate forms it does absorb the constant :D

  5. TuringTest
    • 5 years ago
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    right^

  6. Akshay_Budhkar
    • 5 years ago
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    turing i need u 1 min

  7. Akshay_Budhkar
    • 5 years ago
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    i will give u the link

  8. Akshay_Budhkar
    • 5 years ago
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    http://openstudy.com/study#/updates/4f14d598e4b0b9109c920bca

  9. TuringTest
    • 5 years ago
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    I don't even see why they add the 1 for 'restricted values of x' in wolf what does that mean, do you know akshay?

  10. myininaya
    • 5 years ago
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    what are you guys doing?

  11. myininaya
    • 5 years ago
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    comparing your answers with wolfram?

  12. myininaya
    • 5 years ago
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    sprusty 1/4+constant is still a constant so your answer and the book answer are fine! :)

  13. TuringTest
    • 5 years ago
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    yes, but why does wolf add the 1 for 'restricted values' I want to know

  14. myininaya
    • 5 years ago
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    \[\int\limits_{}^{}\frac{x}{2x+1} dx\] u=2x+1 => du=2 dx u=2x+1 => (u-1)/2=x \[\int\limits_{}^{}x \cdot \frac{1}{2x+1} dx=\int\limits_{}^{}\frac{u-1}{2} \cdot \frac{1}{u} \frac{du}{2}\] \[\frac{1}{4}\int\limits_{}^{}\frac{u-1}{u} du=\frac{1}{4}\int\limits_{}^{}(1-\frac{1}{u}) du=\frac{1}{4}(u-\ln|u|)+C\]

  15. myininaya
    • 5 years ago
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    \[=\frac{1}{4}(2x+1-\ln|2x+1|)+C\]

  16. myininaya
    • 5 years ago
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    \[=\frac{x}{2}+\frac{1}{4}-\frac{1}{4} \ln|2x+1| +C\]

  17. myininaya
    • 5 years ago
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    \[=\frac{x}{2}-\frac{1}{4}\ln|2x+1|+c \]

  18. myininaya
    • 5 years ago
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    i will look at wolfram's solution one sec

  19. myininaya
    • 5 years ago
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    No I don't know what they are talking about honestly lol

  20. myininaya
    • 5 years ago
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    either one of those things they have listed is fine though

  21. myininaya
    • 5 years ago
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    imperialist seems smart I will ask him what he think wolfram means

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