Question

whats the integration of this function?

Answer 1

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

Answer 2

Have you learned about U-Substitution yet?

Answer 3

uhmm nope i dont ! :/

Answer 4

Hm well i can tell you how to do it using U-Substitution but i won't make much sense.

Answer 5

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? (?)

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

divide the du function across by 2:

Answer 11

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

Answer 12

Then your new integral would be:

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

And there you have your answer

Answer 18

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

Answer 19

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.

Answer 20

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