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El_Arrow
 one year ago
need help with integral problem
El_Arrow
 one year ago
need help with integral problem

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's the question/problem

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{?}^{?}\frac{ 3x^33x^2+6x6}{ (x^2+2) (x2)^3}\]

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0there is the integral and i am using partial fraction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Before you do partial fractions, factor the numerator. Nice stuff happens :)

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0this is what i have so far dw:1433371363783:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It will dumb down to something better if you can factor the numerator properly.

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so 3(x^3  x^2 +2x  2)

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0\[x^3  x^2 + 2x2 ~~~=~~~ x^2(x1) + 2(x1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, the numerator is a factoring by grouping example.

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so after that what do i do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, \(x^{2}(x1) + 2(x1)\) can factor also.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, there is a common (x1) term. So factor out the (x1) \(x^{2}(x1) + 2(x1) = (x^{2} + 2)(x1)\) You see how that works?

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so its 3(x^2+2)(x1) right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, which means the \(x^{2} + 2\) would cancel

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0which leaves you with dw:1433371872856:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep. And we can actually avoid partial fractions with that. Because yeah, partial fractions are annoying. \[\int\limits_{}^{}\frac{3x3}{(x2)^{3}}dx = \int\limits_{}^{}\frac{ 3x3 3 + 3 }{ (x2)^{3} }dx =\int\limits_{}^{}\frac{ 3x6 }{ (x2)^{3} }dx+\int\limits_{}^{}\frac{ 3 }{ (x2)^{3} }dx\] See what I did there?

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0how did you get 3+3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I distributed the 3 first, obviously. After, I added and subtracted 3 at the same time, which is equivalent to adding 0, so nothing wrong with that. Doing that allows me to split up the numerator. 3x 3 3 + 3 = 3x6 + 3 Once I get 3x6 + 3, I split those into two separate integrals. The 3x6 is nice because from there I have: \[\int\limits_{}^{}\frac{ 3x6 }{ (x2)^{3} }dx + \int\limits_{}^{}\frac{ 3 }{ (x2)^{3} }dx = 3\int\limits_{}^{}\frac{ (x2) }{ (x2)^{3} }dx + 3\int\limits_{}^{}\frac{ 1 }{ (x2)^{3} }dx\] The + 3 3 thing is just a neat trick that lets you split up fractions faster. Its basically doing partial fractions but without al the long work.

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so if i did partial fractions i would get the same result?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like for example, if I had this integral: \[\int\limits_{}^{}\frac{ 1 }{ e^{x} + 1 }dx\] I would do that same trick of adding and subtracting the same thing. I would do this: \[\int\limits_{}^{}\frac{1+e^{x}  e^{x}}{e^{x}+1}dx = \int\limits_{}^{}\frac{ 1+e^{x} }{ 1+e^{x} }dx  \int\limits_{}^{}\frac{ e^{x} }{ e^{x}+1 }dx\] So it's a good trick to know for integrals. And yes, it would give you the exact same thing.

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0okay but than how would you integrate that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, our integral we have: \[3\int\limits_{}^{}\frac{ 1 }{ (x2)^{2} }dx + 3\int\limits_{}^{}\frac{ 1 }{ (x2)^{3} }dx\] I just canceled the x2 in the first integral. But each of these are just u substitution. Let u = x2

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0quick question if i did use partial fractions i would put the 3 outside the integral right

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0like this?dw:1433372913732:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, putting the 3 outside is preference. I personally dont do it when Im integrating or doing partial fractions or anything, but Ive noticed that a lot of people like moving it outside. I usually keep the 3 in.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So if you wished to do partial fractions, I would just have: \[\frac{ 3x3 }{ (x2)^{3} } = \frac{ A }{ x2 }+ \frac{ B }{ (x2)^{2} } + \frac{ C }{ (x2)^{3} }\]

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0right?dw:1433373114890:dw

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so i put 2 for x right?

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0for C i got 3 and i dont know what number to put for A or B

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, C = 3. Since C = 3, we can do this: \(3x3 = A(x2)^{2} + B(x2) + 3\) \(3x6 = A(x2)^{2}+ B(x2)\) From here, you have the option of distributing everything out and equating powers or we can pick a couple random values of x and solve a system of equations.

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0which option is easier?

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0which option is easier?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, foiling out is pretty easy, so I would just do that. If the foiling was annoying, I might do the system route. But you can foil this out and get: \(3x6 = A(x^{2} 4x + 4) + Bx  2B\) \(3x6 = Ax^{2}  4Ax + 4A + Bx  2B\) So by equating powers, the left has no \(x^{2}\) terms and the right side has \(Ax^{2}\). This means A has to be 0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since A is 0, you then would have: \(3x6 = B(x2)\) Pick x to be 0 and you get \(6 = 2B\) so B = 3, which is what we had through the tricky method, we got 3 and 3.
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