PARTIAL FRACTIONS example with distinct linear factors

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Krista King

Krista King

Күн бұрын

► My Integrals course: www.kristaking...
The tricky thing about partial fractions is that there are four kinds of partial fractions problems. The kind of partial fractions decomposition you'll need to perform depends on the kinds of factors in your denominator.
You can have 1) Distinct linear factors, 2) Distinct quadratic factors, 3) Repeated linear factors, or 4) Repeated quadratic factors. You can also have a mix of any and all of these.
In this video, we're going to look at a partial fractions problem with distinct linear factors. We'll do the partial fractions decomposition, and then use the method of undetermined coefficients to solve for our constants. Then we'll plug these back into the decomposition, and the decomposition back into the integral, and then we'll be able to evaluate the integral.
Remember, partial fractions in general is a method of integration that you can use to simplify rational functions to make them easier to integrate. If it's possible to integrate the function directly, or to use u-substitution or integration by parts, those methods will all likely be easier and quicker than partial fractions, so you should consider them before you try this.
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Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”
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Пікірлер: 8
@muhammadhamzajaved9297
@muhammadhamzajaved9297 8 жыл бұрын
A great example and so refreshing after waking up from sleep and watching this video 😀 . Great work 👏
@kristakingmath
@kristakingmath 8 жыл бұрын
Thanks, Hamza!
@ahmedaldakhoul5093
@ahmedaldakhoul5093 8 жыл бұрын
amazing Krista 👍👍
@kristakingmath
@kristakingmath 8 жыл бұрын
Thank you so much, Ahmed!
@jamesbond_007
@jamesbond_007 8 жыл бұрын
Great explanation - thanks. I'm a little confused -- maybe my understanding is incorrect: I thought the requirement for these partial fractions problems was that the degree of the numerator had to be less than that of the highest degree of the denominator factors, yet here it is not the case. Now, it works here because any 2 of the 3 denominator factors make a second degree polynomial, but still...am I misremembering the rule, or is this a special case?
@TheNetkrot
@TheNetkrot 5 жыл бұрын
great ...thanks
@muhammadhamzajaved9297
@muhammadhamzajaved9297 8 жыл бұрын
So whats next for you Miss? My suggestion is that you work on conics like circles ,parabolas ,ellipse and hyperbola ...... 😊 Cuz these things has too many concepts and formulae that is sometimes difficult to grasp and understand ..
@luisl.b4146
@luisl.b4146 8 жыл бұрын
Wow.
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