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whats the integration of this function?

Mathematics
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\[\ \int\limits_{}^{} \frac{ 2x }{ 2x ^{2}+1 }\]
Have you learned about U-Substitution yet?
uhmm nope i dont ! :/

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Other answers:

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

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