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appleduardo

  • 2 years ago

whats the integration of this function?

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  1. appleduardo
    • 2 years ago
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    \[\ \int\limits_{}^{} \frac{ 2x }{ 2x ^{2}+1 }\]

  2. sanchez9457
    • 2 years ago
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    Have you learned about U-Substitution yet?

  3. appleduardo
    • 2 years ago
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    uhmm nope i dont ! :/

  4. sanchez9457
    • 2 years ago
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    Hm well i can tell you how to do it using U-Substitution but i won't make much sense.

  5. appleduardo
    • 2 years ago
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    i just searched what u substitution means, and i noticed that i cant make use of formulas.. in this case may i use the S du/u = ln u +c? (?)

  6. sanchez9457
    • 2 years ago
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    well you could set 2x^2 to be your u

  7. sanchez9457
    • 2 years ago
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    when you derive that you get a simple function you could use.

  8. sanchez9457
    • 2 years ago
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    sorry i meant set your u to be 2x^2 + 1

  9. sanchez9457
    • 2 years ago
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    \[ u = 2x^2 +1\] \[du = 4xdx\]

  10. sanchez9457
    • 2 years ago
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    divide the du function across by 2:

  11. sanchez9457
    • 2 years ago
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    \[\frac{ du }{ 2 } = 2xdx\]

  12. sanchez9457
    • 2 years ago
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    Then your new integral would be:

  13. sanchez9457
    • 2 years ago
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    \[\frac{ 1 }{ 2 }\int\limits_{?}^{?} \frac{ 1 }{ u } du\]

  14. sanchez9457
    • 2 years ago
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    if you integrate that you get: \[\frac{ 1 }{ 2 }\left[ logu \right] + C\]

  15. sanchez9457
    • 2 years ago
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    then replace that u with the equation it was set to:

  16. sanchez9457
    • 2 years ago
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    \[= \frac{ 1 }{ 2 } \log(2x^2 +1)\]

  17. sanchez9457
    • 2 years ago
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    And there you have your answer

  18. appleduardo
    • 2 years ago
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    oh my gosh thank you so much ! . :D but i have a question, why did u divided du by 2?

  19. sanchez9457
    • 2 years ago
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    i divided du by 2 so it can look like something i already had. I could have done it withoug dividing by 2 but i like to simplify things.

  20. sanchez9457
    • 2 years ago
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    by somthign i had already i meant 2x. from the original integral.

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