appleduardo
whats the integration of this function?



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appleduardo
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\[\ \int\limits_{}^{} \frac{ 2x }{ 2x ^{2}+1 }\]

sanchez9457
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Have you learned about USubstitution yet?

appleduardo
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uhmm nope i dont ! :/

sanchez9457
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Hm well i can tell you how to do it using USubstitution but i won't make much sense.

appleduardo
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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? (?)

sanchez9457
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well you could set 2x^2 to be your u

sanchez9457
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when you derive that you get a simple function you could use.

sanchez9457
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sorry i meant set your u to be 2x^2 + 1

sanchez9457
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\[ u = 2x^2 +1\]
\[du = 4xdx\]

sanchez9457
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divide the du function across by 2:

sanchez9457
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\[\frac{ du }{ 2 } = 2xdx\]

sanchez9457
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Then your new integral would be:

sanchez9457
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\[\frac{ 1 }{ 2 }\int\limits_{?}^{?} \frac{ 1 }{ u } du\]

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

sanchez9457
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then replace that u with the equation it was set to:

sanchez9457
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\[= \frac{ 1 }{ 2 } \log(2x^2 +1)\]

sanchez9457
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And there you have your answer

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

sanchez9457
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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.

sanchez9457
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by somthign i had already i meant 2x. from the original integral.