appleduardo
  • appleduardo
whats the integration of this function?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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appleduardo
  • appleduardo
\[\ \int\limits_{}^{} \frac{ 2x }{ 2x ^{2}+1 }\]
sanchez9457
  • sanchez9457
Have you learned about U-Substitution yet?
appleduardo
  • appleduardo
uhmm nope i dont ! :/

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sanchez9457
  • sanchez9457
Hm well i can tell you how to do it using U-Substitution but i won't make much sense.
appleduardo
  • appleduardo
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? (?)
sanchez9457
  • sanchez9457
well you could set 2x^2 to be your u
sanchez9457
  • sanchez9457
when you derive that you get a simple function you could use.
sanchez9457
  • sanchez9457
sorry i meant set your u to be 2x^2 + 1
sanchez9457
  • sanchez9457
\[ u = 2x^2 +1\] \[du = 4xdx\]
sanchez9457
  • sanchez9457
divide the du function across by 2:
sanchez9457
  • sanchez9457
\[\frac{ du }{ 2 } = 2xdx\]
sanchez9457
  • sanchez9457
Then your new integral would be:
sanchez9457
  • sanchez9457
\[\frac{ 1 }{ 2 }\int\limits_{?}^{?} \frac{ 1 }{ u } du\]
sanchez9457
  • sanchez9457
if you integrate that you get: \[\frac{ 1 }{ 2 }\left[ logu \right] + C\]
sanchez9457
  • sanchez9457
then replace that u with the equation it was set to:
sanchez9457
  • sanchez9457
\[= \frac{ 1 }{ 2 } \log(2x^2 +1)\]
sanchez9457
  • sanchez9457
And there you have your answer
appleduardo
  • appleduardo
oh my gosh thank you so much ! . :D but i have a question, why did u divided du by 2?
sanchez9457
  • sanchez9457
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.
sanchez9457
  • sanchez9457
by somthign i had already i meant 2x. from the original integral.

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