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anonymous

  • 5 years ago

if I have an expression like this x/(x+1). I can divide numerator by denominator and get something like this 1-1/(x+1). Can someone teach me how to ...

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  1. anonymous
    • 5 years ago
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    express the numerator as \[\frac{ x+1-1}{x+1} = \frac{ x+1}{x+1} + \frac{-1}{x+1} = 1 \frac{1}{x+1} \]

  2. anonymous
    • 5 years ago
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    \[1- \frac{1}{x+1}\] that should be last part

  3. anonymous
    • 5 years ago
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    we add and subtract the same thing, so we dont change the value overall

  4. anonymous
    • 5 years ago
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    can you give me another example?

  5. anonymous
    • 5 years ago
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    take \[\frac{x+2}{x+3} \]

  6. anonymous
    • 5 years ago
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    so we look on the bottom , it has x+3 , so we want to see an x+3 in the numerator

  7. anonymous
    • 5 years ago
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    so express as \[\frac{x+3+2 -3 }{x+3} = \frac{x+3 -1 }{x+3}\]

  8. anonymous
    • 5 years ago
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    \[= \frac{x+3}{x+3} - \frac{1}{x+3} = 1 - \frac{1}{x+3}\]

  9. anonymous
    • 5 years ago
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    its all fairly standard

  10. anonymous
    • 5 years ago
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    so you mean I have to look at the bottom. if I have for example (x+4), then I have to add and substract 4 and -4 respectively?

  11. anonymous
    • 5 years ago
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    yes , all it relys on is being able to split up the numerators

  12. anonymous
    • 5 years ago
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    in the numerator

  13. anonymous
    • 5 years ago
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    ok im gonna give you an example. Please check if it is correct

  14. anonymous
    • 5 years ago
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    \[x/(1+x^{2})\]well i dont know what to do here :(

  15. anonymous
    • 5 years ago
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    thats why it only works with linear terms in the numerator and denominator

  16. anonymous
    • 5 years ago
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    you cant break that up

  17. anonymous
    • 5 years ago
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    partial fractions? idk you need a constant on top

  18. anonymous
    • 5 years ago
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    but that cant be broken down

  19. anonymous
    • 5 years ago
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    ya

  20. anonymous
    • 5 years ago
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    you cant just make up random examples

  21. anonymous
    • 5 years ago
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    or you could have something like \[\frac{x^2 +2x} {(x+1)^2} \]

  22. anonymous
    • 5 years ago
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    So how should I do in cases like this. that is why I dont understand The oiginal problems starts there

  23. anonymous
    • 5 years ago
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    the question is badly written, I would just ignore it, it cant be broken down further , teacher is stupid

  24. anonymous
    • 5 years ago
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    Sorry dude but im working with integrals. They have an expression like that and get that answer. the answer is 1-1(x2 +1)

  25. anonymous
    • 5 years ago
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    I dont understand whart u mean

  26. anonymous
    • 5 years ago
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    but if you want \[\int\limits \frac{x}{1+x^2} dx \]

  27. anonymous
    • 5 years ago
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    then its a logarithm

  28. anonymous
    • 5 years ago
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    yeah that is right

  29. anonymous
    • 5 years ago
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    in a book im reading, they say that \[x ^{2}/(1+x ^{2})=1-1/(1+x ^{2})\]

  30. anonymous
    • 5 years ago
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    \[= \frac{1}{2} \ln ( 1+x^2) +\]

  31. anonymous
    • 5 years ago
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    my question is how do I do that

  32. anonymous
    • 5 years ago
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    ohh yeh, now you have changed the question :|

  33. anonymous
    • 5 years ago
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    sorry lol

  34. anonymous
    • 5 years ago
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    add and subtract 1 on the top and bottom

  35. anonymous
    • 5 years ago
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    \[=\frac{x^2+1-1}{x^2+1} = \frac{x^2+1}{x^2+1} - \frac{1}{x^2+1} = 1- \frac{1}{x^2+1}\]

  36. anonymous
    • 5 years ago
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    very basic, nothing really hard

  37. anonymous
    • 5 years ago
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    nice

  38. anonymous
    • 5 years ago
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    ok cool. sorry for me to make you feel angry lol

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