Integrate
\(\frac{-\frac{1}{\sqrt{2}v}}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}\)

- inkyvoyd

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- schrodinger

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- inkyvoyd

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

- inkyvoyd

Wait, correction.

- inkyvoyd

\(\frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}\)

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## More answers

- inkyvoyd

Am I supposed to complete the square?

- inkyvoyd

@dpaInc , pwease help me!

- anonymous

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

- inkyvoyd

@dpaInc , how do you copy it like that?

- anonymous

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

- anonymous

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

- inkyvoyd

Alright, now solve NAO!

- inkyvoyd

xD

- inkyvoyd

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

- anonymous

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

- inkyvoyd

lolol

- inkyvoyd

@dpaInc , now do I complete the square?

- anonymous

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

- anonymous

there it is.... #13....:)
http://integral-table.com/downloads/single-page-integral-table.pdf

- anonymous

i mean #16...l

- inkyvoyd

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

- inkyvoyd

But isn't using tables cheating?

- anonymous

no... of course not...

- inkyvoyd

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

- anonymous

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?

- inkyvoyd

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

- anonymous

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

- inkyvoyd

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

- anonymous

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...

- inkyvoyd

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

- anonymous

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

- inkyvoyd

I already did in my last tex post :)

- 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

- anonymous

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

- inkyvoyd

xD

- anonymous

you are solving the problem in a wrong way

- inkyvoyd

@nitz , what am I doing wrong?

- anonymous

see firstly make numerator as the derivative of denominator

- inkyvoyd

Can you show me?

- anonymous

i am talking about 1st integration ie before addition

- inkyvoyd

?

- anonymous

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

- inkyvoyd

You mean at the very start of the problem?

- anonymous

ya

- inkyvoyd

but but but

- 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 >.<

- anonymous

you cant solve it further in this case

- inkyvoyd

#16?
http://integral-table.com/downloads/single-page-integral-table.pdf

- anonymous

these are direct problems

- anonymous

these are direct solutions

- anonymous

you can apply direct formula

- anonymous

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

- inkyvoyd

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

- anonymous

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.

- anonymous

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

- anonymous

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.

- inkyvoyd

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

- anonymous

nbouscal: Integral Detective

- inkyvoyd

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

- anonymous

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.

- anonymous

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

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