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

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

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Other answers:

Am I supposed to complete the square?
@dpaInc , pwease help me!
can't see it so i'm gonna copy it... and make it huge... \[\huge \frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1} \]
@dpaInc , how do you copy it like that?
right click-->show math as-->tex commands
\[\huge \int\frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}dv \]
Alright, now solve NAO!
btw, this is the equation I got after trying 3 substitutions and seperating the fraction.
\[\large \int\frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}dv\] \[\large \frac{-1}{\sqrt2}\int\frac{v}{v^2+\sqrt{2}v+1}dv+\frac{1}{\sqrt2}\int \frac{v}{v^2-\sqrt{2}v+1}dv\] maybe i should just draw...
@dpaInc , now do I complete the square?
yeah... that's what i was about to do... but i think there is a formula... hang on....
there it is.... #13....:)
i mean #16...l
\(\huge \frac{-1}{\sqrt2}\int\frac{v}{(v+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv+\frac{1}{\sqrt2}\int \frac{v}{(v-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv\)
But isn't using tables cheating?
no... of course not...
But... I don't know where the tables come from... >.<
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?
i don't think it's cheating... it's a good shortcut to cut through all that stuff..
Okay... @dpaInc , without showing me the math, can you tell me what to do after I have completed the square?
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...
@dpaInc , my problem is that I'm not sure how to get the results shown in the tables.
let me complete the square on the first integral... i haven't done it for a while and i guess i need the practice...
I already did in my last tex post :)
\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
oh... i see you did it already....
you are solving the problem in a wrong way
@nitz , what am I doing wrong?
see firstly make numerator as the derivative of denominator
Can you show me?
i am talking about 1st integration ie before addition
in it, firstly multiply the numerator with 2 and add and subtract \[\sqrt{ 2} \] in it
You mean at the very start of the problem?
but but but
\(\huge \frac{-1}{\sqrt2}\int\frac{v}{(v+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv+\frac{1}{\sqrt2}\int \frac{v}{(v-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv\) we already got to here >.<
you cant solve it further in this case
these are direct problems
these are direct solutions
you can apply direct formula
i think it works.... |dw:1340004704215:dw|
Uh, I can't use u in substiution LOL. I already used it...
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.
Next OS question: What made this integral appealing? I will medal the person with the best response.
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.
@nbouscal , that was my integration of the square root of tan x...
nbouscal: Integral Detective
Maybe I should show what work i did before...
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.

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