El_Arrow
  • El_Arrow
need help with integral problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
?
anonymous
  • anonymous
What's the question/problem
El_Arrow
  • El_Arrow
i am typing it man

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anonymous
  • anonymous
ok
El_Arrow
  • El_Arrow
\[\int\limits_{?}^{?}\frac{ 3x^3-3x^2+6x-6}{ (x^2+2) (x-2)^3}\]
El_Arrow
  • El_Arrow
there is the integral and i am using partial fraction
anonymous
  • anonymous
Before you do partial fractions, factor the numerator. Nice stuff happens :)
El_Arrow
  • El_Arrow
this is what i have so far |dw:1433371363783:dw|
anonymous
  • anonymous
It will dumb down to something better if you can factor the numerator properly.
El_Arrow
  • El_Arrow
so 3(x^3 - x^2 +2x - 2)
geerky42
  • geerky42
Factor more.
geerky42
  • geerky42
\[x^3 - x^2 + 2x-2 ~~~=~~~ x^2(x-1) + 2(x-1)\]
El_Arrow
  • El_Arrow
oh
anonymous
  • anonymous
Yeah, the numerator is a factoring by grouping example.
El_Arrow
  • El_Arrow
so after that what do i do
anonymous
  • anonymous
Well, \(x^{2}(x-1) + 2(x-1)\) can factor also.
El_Arrow
  • El_Arrow
in to what?
anonymous
  • anonymous
Well, there is a common (x-1) term. So factor out the (x-1) \(x^{2}(x-1) + 2(x-1) = (x^{2} + 2)(x-1)\) You see how that works?
El_Arrow
  • El_Arrow
so its 3(x^2+2)(x-1) right
anonymous
  • anonymous
Yep, which means the \(x^{2} + 2\) would cancel
El_Arrow
  • El_Arrow
which leaves you with |dw:1433371872856:dw|
anonymous
  • anonymous
Yep. And we can actually avoid partial fractions with that. Because yeah, partial fractions are annoying. \[\int\limits_{}^{}\frac{3x-3}{(x-2)^{3}}dx = \int\limits_{}^{}\frac{ 3x-3 -3 + 3 }{ (x-2)^{3} }dx =\int\limits_{}^{}\frac{ 3x-6 }{ (x-2)^{3} }dx+\int\limits_{}^{}\frac{ 3 }{ (x-2)^{3} }dx\] See what I did there?
El_Arrow
  • El_Arrow
no i am lost
El_Arrow
  • El_Arrow
how did you get -3+3?
anonymous
  • anonymous
I 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 = 3x-6 + 3 Once I get 3x-6 + 3, I split those into two separate integrals. The 3x-6 is nice because from there I have: \[\int\limits_{}^{}\frac{ 3x-6 }{ (x-2)^{3} }dx + \int\limits_{}^{}\frac{ 3 }{ (x-2)^{3} }dx = 3\int\limits_{}^{}\frac{ (x-2) }{ (x-2)^{3} }dx + 3\int\limits_{}^{}\frac{ 1 }{ (x-2)^{3} }dx\] 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
  • El_Arrow
so if i did partial fractions i would get the same result?
anonymous
  • anonymous
Like 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
  • El_Arrow
okay but than how would you integrate that?
anonymous
  • anonymous
Well, our integral we have: \[3\int\limits_{}^{}\frac{ 1 }{ (x-2)^{2} }dx + 3\int\limits_{}^{}\frac{ 1 }{ (x-2)^{3} }dx\] I just canceled the x-2 in the first integral. But each of these are just u substitution. Let u = x-2
El_Arrow
  • El_Arrow
quick question if i did use partial fractions i would put the 3 outside the integral right
El_Arrow
  • El_Arrow
like this?|dw:1433372913732:dw|
anonymous
  • anonymous
Well, 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
  • anonymous
So if you wished to do partial fractions, I would just have: \[\frac{ 3x-3 }{ (x-2)^{3} } = \frac{ A }{ x-2 }+ \frac{ B }{ (x-2)^{2} } + \frac{ C }{ (x-2)^{3} }\]
El_Arrow
  • El_Arrow
ok i see
El_Arrow
  • El_Arrow
right?|dw:1433373114890:dw|
anonymous
  • anonymous
Yes
El_Arrow
  • El_Arrow
so i put 2 for x right?
El_Arrow
  • El_Arrow
for C i got 3 and i dont know what number to put for A or B
anonymous
  • anonymous
Yeah, C = 3. Since C = 3, we can do this: \(3x-3 = A(x-2)^{2} + B(x-2) + 3\) \(3x-6 = A(x-2)^{2}+ B(x-2)\) 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
  • El_Arrow
which option is easier?
El_Arrow
  • El_Arrow
which option is easier?
anonymous
  • anonymous
Well, 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: \(3x-6 = A(x^{2} -4x + 4) + Bx - 2B\) \(3x-6 = Ax^{2} - 4Ax + 4A + Bx - 2B\) 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
  • anonymous
Since A is 0, you then would have: \(3x-6 = B(x-2)\) 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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