Integration Using method of Partial Fractions
Integration Using method of Partial Fractions - A complete example! For more free math videos, visit http://JustMathTutoring.com! I have organized links to over 100 FREE math videos made by me!
Duration : 0:9:50
[youtube 6qVgHWxdlZ0]
Tags: calculus, coefficients, decomposition, equate, example, fractions, integrate, integration, justmathtutoring.com, math, partial
July 28th, 2009 at 4:42 am
please be careful …
please be careful with your parenthesis..
July 28th, 2009 at 4:42 am
how do you know to …
how do you know to plug in 0?
July 28th, 2009 at 4:42 am
i dont mean to …
i dont mean to sound rude but not putting Dx+E in the equation makes the rest of the video hard to watch. why not a redo? mulligan?
July 28th, 2009 at 4:42 am
ehh so…. in the …
ehh so…. in the beginning, why do we need to have the a
A/x term when there’s really no single x value in the original equation?
July 28th, 2009 at 4:42 am
ha.
i often find …
ha.
i often find myself disliking humanity so much, i wonder why i make these, and wonder if one day i will just take them all down… : )
July 28th, 2009 at 4:42 am
it’s people like …
it’s people like you that give me hope for humanity
July 28th, 2009 at 4:42 am
Because there is (1 …
)
Because there is (1+x^2). The problem is the x^2. When you’ve got x^2, you must write Dx+E (not always Dx+E, of course - it can be Gx+H
July 28th, 2009 at 4:42 am
why isnt it just D?
why isnt it just D?
July 28th, 2009 at 4:42 am
yes, absolutely.
a …
yes, absolutely.
a few others pointed this out; i have added annotations pointing out the mistake.
thanks for pointing it out!!
July 28th, 2009 at 4:42 am
why is there an d …
why is there an d over 1+x^2 it should be (dx+e)
July 28th, 2009 at 4:42 am
Where Can I find …
Where Can I find the corrected video?
July 28th, 2009 at 4:42 am
… I did it all …
… I did it all without the “trick” at the start where you set x=0 and x=-1, where I just multiplied it all out and solved using equating of the coefficients, and after all was done, I got “1/x^2+-1/x”. Did I carry this out correctly? Would you have gotten the same if you had multiplied out completely / didn’t substitute in your values that you got from your “trick”?
July 28th, 2009 at 4:42 am
thanks this helped …
thanks this helped a lot…i was wondering though, why do you have to use arctan for the last term, my professor is horrible at explaining things…thanks
July 28th, 2009 at 4:42 am
ops yes,i will have …
ops yes,i will have to fix this!
July 28th, 2009 at 4:42 am
I agree with what …
I agree with what you are saying.
July 28th, 2009 at 4:42 am
Great video, thanks …
Great video, thanks very much
July 28th, 2009 at 4:42 am
Are you sure the …
Are you sure the last term isn’t Dx + E in the numerator? Usually if (b^2 - 4ac) for the factor is negative, you use Dx + E (or whatever letters/variables your using) rather than just a D.
(0^2 - 4(1)(1)) = -4 which is < 0. For (x^2+1)
July 28th, 2009 at 4:42 am
that is the partial …
that is the partial fraction decomposition for linear factors… i have a video on my website explaining that step if you feel the need to check it out!
July 28th, 2009 at 4:42 am
how did you get A …
how did you get A over x when you were starting off?
July 28th, 2009 at 4:42 am
yea, partial …
yea, partial fractions is a long procedure unfortunately! hope it makes sense though!
July 28th, 2009 at 4:42 am
whew, long but good!
whew, long but good!