## inkyvoyd Integrate $$\frac{-\frac{1}{\sqrt{2}v}}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}$$ one year ago one year ago

1. inkyvoyd

@nbouscal will know what I am talking about, or maybe @Limitless

2. inkyvoyd

Wait, correction.

3. inkyvoyd

$$\frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}$$

4. inkyvoyd

Am I supposed to complete the square?

5. inkyvoyd

@dpaInc , pwease help me!

6. dpaInc

can't see it so i'm gonna copy it... and make it huge... $\huge \frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}$

7. inkyvoyd

@dpaInc , how do you copy it like that?

8. dpaInc

right click-->show math as-->tex commands

9. dpaInc

$\huge \int\frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}dv$

10. inkyvoyd

Alright, now solve NAO!

11. inkyvoyd

xD

12. inkyvoyd

btw, this is the equation I got after trying 3 substitutions and seperating the fraction.

13. dpaInc

$\large \int\frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}dv$ $\large \frac{-1}{\sqrt2}\int\frac{v}{v^2+\sqrt{2}v+1}dv+\frac{1}{\sqrt2}\int \frac{v}{v^2-\sqrt{2}v+1}dv$ maybe i should just draw...

14. inkyvoyd

lolol

15. inkyvoyd

@dpaInc , now do I complete the square?

16. dpaInc

yeah... that's what i was about to do... but i think there is a formula... hang on....

17. dpaInc

18. dpaInc

i mean #16...l

19. inkyvoyd

$$\huge \frac{-1}{\sqrt2}\int\frac{v}{(v+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv+\frac{1}{\sqrt2}\int \frac{v}{(v-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv$$

20. inkyvoyd

But isn't using tables cheating?

21. dpaInc

no... of course not...

22. inkyvoyd

But... I don't know where the tables come from... >.<

23. dpaInc

it's like using all the different tires for your car... you're not going to make/invent your own tires are you? oh wait.. do you drive yet?

24. inkyvoyd

No lol. I'M GOING TO BUILD MY CAR FROM SCRATCH MUAHAHAHA

25. dpaInc

i don't think it's cheating... it's a good shortcut to cut through all that stuff..

26. inkyvoyd

Okay... @dpaInc , without showing me the math, can you tell me what to do after I have completed the square?

27. dpaInc

and as for me, i haven't really started using the integral tables until i saw everyon virtually used them here on OS.... i actually went through proving stuff because I don't use/memorize the tables...

28. inkyvoyd

@dpaInc , my problem is that I'm not sure how to get the results shown in the tables.

29. dpaInc

let me complete the square on the first integral... i haven't done it for a while and i guess i need the practice...

30. inkyvoyd

I already did in my last tex post :)

31. inkyvoyd

\huge \frac{-1}{\sqrt2}\int\frac{v}{(v+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv+\frac{1}{\sqrt2}\int \frac{v}{(v-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv

32. dpaInc

oh... i see you did it already....

33. inkyvoyd

xD

34. nitz

you are solving the problem in a wrong way

35. inkyvoyd

@nitz , what am I doing wrong?

36. nitz

see firstly make numerator as the derivative of denominator

37. inkyvoyd

Can you show me?

38. nitz

39. inkyvoyd

?

40. nitz

in it, firstly multiply the numerator with 2 and add and subtract $\sqrt{ 2}$ in it

41. inkyvoyd

You mean at the very start of the problem?

42. nitz

ya

43. inkyvoyd

but but but

44. inkyvoyd

$$\huge \frac{-1}{\sqrt2}\int\frac{v}{(v+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv+\frac{1}{\sqrt2}\int \frac{v}{(v-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv$$ we already got to here >.<

45. nitz

you cant solve it further in this case

46. inkyvoyd
47. nitz

these are direct problems

48. nitz

these are direct solutions

49. nitz

you can apply direct formula

50. dpaInc

i think it works.... |dw:1340004704215:dw|

51. inkyvoyd

Uh, I can't use u in substiution LOL. I already used it...

52. nbouscal

Okay, I'm late to this party, but you can just complete the square and then do a substitution. This is just like the case that we reach in the integration of sqrt(tan x) that we have already done, inky.

53. Limitless

Next OS question: What made this integral appealing? I will medal the person with the best response.

54. nbouscal

This actually looks very similar to the integral of sqrt(tan x), if I had to guess I would say it is an intermediate step in a similar integration.

55. inkyvoyd

@nbouscal , that was my integration of the square root of tan x...

56. Limitless

nbouscal: Integral Detective

57. inkyvoyd

Maybe I should show what work i did before...

58. nbouscal

I didn't actually go back and look at that integral, you're probably right on. The next step is indeed to complete the square, then do another substitution.

59. nbouscal

Oh hey just look here: http://openstudy.com/study#/updates/4fb3c611e4b0556534298c6b