Integrate 1 / (x2 + 3x + 2)

I dont know anything about calculus im not full of shit at all. I just want to see how this tough sucker is solved

1/(x+1)(x+2)
1/(x+1) - 1/(x+2)
ln|x+1| - ln|x+2| + K

okay so it's just a way to express ideas with some symbols that are sort of different from english

so a set is a collection of objects. can be anything - numbers, oranges, orangutans, ideas, whatever.

so what i'm doing here is defining a set. i'm giving it a name, which is R.

the semantics of it are the curly brackets, the way i wrote the sentence, the nitty gritty. so what i said was...

R = x. x is a variable that i'm defining in this sentence and it's a set, denoted by the curly brackets.

so R equals all of x. then there's the pipe / bar / whatever which tells you what x is all about

the backwards e with the slash symbol means "is not an element of". so what it's saying is that for every x, it's not a part of whatever set i define after the backwards e thingy.

the parentheses mean - more or less - that i'm defining a function-variable... whatever is going on within the parentheses has to be evaluated to make sense of the rest of what's going on

within the parentheses, i've just clearly defined... fucks fat chicks. so x, not being an element of the set that fucks fat chicks, returns to the set definition that R is every element that doesn't fuck fat chicks, and is that set.

what i'm trying to say is no fatties.

1 / (x2 + 3x + 2)

1 / (x + 2)(x + 1)

1/(x+2) * 1/(x+1)

n^(x+1) * n^(x)

n^2x(x+1)

n^(2x^2+2)

1/(2x^2+3)

...

Repulsive.

Honestly though, I'm jealous of people who can live with being a chubby chaser. I know this guy from my department who has a fat chick as a gf. He looks pretty normal, not great, just okay. I know everybody secretly looks down on him. It's just this thing you keep in mind, it kind of limits the respect you can have for him. But overall, must be amazing to just run around with a disgusting pig like that, for real. I would be so ashamed.

I love fucking fat chicks right in their big fat fucking asses with my massive fat-fucking cock

Okay so I don't want to get into number theory here. But we can still talk about sets.

So the girls you posted are represented by the variable G. Let's say that R is now the set of girls who exist. So what you just posted is
G (1, 2) where G = (x | x ∉ R)
so these two chicks don't exist IRL and are photoshopped to have impossibly huge asses.
That's chill, but then we can define a set of F which is girls than you can fuck. We could use a lambda for this to define the function of fucking these girls, but it's kind of pointless. Easier to just be like
F e R,
F e G
which we already defined as the set of girls that are real. These chicks ain't real. And there's plenty of 2 hundos that I have, and would, stick my dong into, but that shit gets complicated because it involved differentials. Pretty much what I'm getting at is yes, I would and have fucked fat chicks, but none of them look like that photoshopped shit and get real son.

Neither of these responses are part of the set of things I responded to you with. I'm gonna namefag for a minute to clear up confusion because I love teaching set theory.

So let's get to something interesting... what if you define a set as every set that doesn't include itself?

> 1/(x+1)(x+2) = 1/(x+1) - 1/(x+2)

you can't do that you nigger, the step inbetween has to be
1/(x+1)(x+2) = A/(x+1) - B/(x+2)

and then you solve A and B, then you can integrate it

...

That's a pawg. Look, just cuz I jack off to type theory videos all day doesn't mean I don't get mad puss. She's probably some kind of spanish. WASPs don't have asses like that without butt implants. Anyway, probably wouldn't smash because she looks high maintenance.

Now represent what I just said in set notation so we can make sure you're learning.

Let's not even get into this now... okay thick white girls

Picky retard.

m o ya

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[math]\int\frac{1}{x^2+3x+2}dx[/math]

[math] =\int\frac{1}{(x+2)(x+1)}dx [/math]

[math] =\int\frac{A}{x+2}+\frac{b}{x+1} dx [/math]

solve for A and B.
Find that [math] A=-1, B=1[/math].

[math] \therefore\int-\frac{1}{x+2}dx+\int \frac{1}{x+1}dx [/math]

[math]\therefore ans= ln|\frac{x+1}{x+2}|+c [/math]

ok

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Okay I now understand that calculus is not real math

Trips also can u name every school od mathematics

Omg what's her name???

nigga how the FUCK you did that

Veeky Forums supports tex this whole time and I was just like too high and autistic to ever bother with it? wtf i thought that was a joke

it's fucking algebra. it's the exact same shit. it's a branch of algebra.

...

oh yeah? how did you get the values of a and b

fucking algebra you retard, how do you get the values of any variable that you defined? it's literally all algebra, integrals & derivatives & differential eqs are algebra. it's all the same shit and it all comes down to number theory

how did you get the value of 1

see, you need some computational theory for that. some lambda calculus. but it's all still algebra, euler's constant is still algebra, might need to fancy it up a little with some set theory but (gasp) it's all algebra, and algebra is just math. it's all the same shit. like geometry too, it's all the same shit. as long as you can define 1 = 1 you can do any math that has ever been invented if you're not a total fucking moron