## appleduardo Group Title whats the integration of this function? one year ago one year ago

1. appleduardo Group Title

$\ \int\limits_{}^{} \frac{ 2x }{ 2x ^{2}+1 }$

2. sanchez9457 Group Title

Have you learned about U-Substitution yet?

3. appleduardo Group Title

uhmm nope i dont ! :/

4. sanchez9457 Group Title

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

5. appleduardo Group Title

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 Group Title

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

7. sanchez9457 Group Title

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

8. sanchez9457 Group Title

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

9. sanchez9457 Group Title

$u = 2x^2 +1$ $du = 4xdx$

10. sanchez9457 Group Title

divide the du function across by 2:

11. sanchez9457 Group Title

$\frac{ du }{ 2 } = 2xdx$

12. sanchez9457 Group Title

Then your new integral would be:

13. sanchez9457 Group Title

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

14. sanchez9457 Group Title

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

15. sanchez9457 Group Title

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

16. sanchez9457 Group Title

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

17. sanchez9457 Group Title

18. appleduardo Group Title

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

19. sanchez9457 Group Title

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 Group Title

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