Antiderivative

Sup Veeky Forums does anyone have a good e book or some shit like that to learn how to find antiderivatives ?
also what the fuck is the anti derivative of (x+1)/(x^2-x+1)
does it even exist ?
I'm going to kill myself I'm losing my mind over this crap

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hey fag go to /adv/ with homework help

also check the sticky retard

fuck you the sticky is too long to read
pls help

>makes thread for help with calc II problems
>expecting anything besides getting told to fuck off

kek

stop acting all smart and shit you nerd
you probably can't even do it

fucking calc ii is hard tho
Ive got only a few weeks til the final and I feel like Ive learned jack shit the entire semester.
send help.

Fuck this board is shit I'll go back to /pol/ no wonder you guys are irrelevant and no one likes you

>WAHHH nobody helped me with my shitty calculus problems after my thread has been up for 10 mins
cry me a river

Do a partial fraction decomposition with the ansatz:

[eqn] \frac{x}{x^2 - x + 1} = \frac{A}{x - e^{\frac{i \pi}{3}}} + \frac{B}{x - e^{\frac{-i \pi}{3}}} [/eqn]

instead of acting all smart and shit and saying that it's simple and all that
why don't you fags help a brotha out ?

Holy mother of god

You need to know three things for sure:
>u substitution
>u-v substitution
>trig substitution
Also, when you have factored polynomials on the denominator, sometimes it makes sense to use partial fractions

I believe there are other tricks, but these are pretty much the main ones.

So basically, we can only truly factor 2 integrals. That is, the integral of e^kx and the integral x^n. All of these other substitutions, tricks, and formulas are just a complicated way to get the equation into the form of either an exponential, or a polynomial. This is because these are the only Riemann sums we can easily factor

So the thing with antiderivatives, is you can prove that a derivative undoes an integral and vice versa. If you want to see this for yourself, plug the formula for an integral of a function into the formula for a derivative, or vice-versa. Therefore, one way to think about an integral is as an anti-derivative (that is, a reverse derivative). So you can consider what type of function would lead to the integrand when a derivative is done.

So in order to manipulate our integrand into a form that can be machined by the integration operator, we use the techniques I outlined above.
>u substitution
Is basically backwards chain rule.
Instead of taking the derivative of f(g(x)) to get f'(g(x))g'(x), you are doing the reverse
>u-v substitution
Is basically backwards product rule.
Instead of taking the derivative of f(x)g(x) to get f'(x)g(x) + f(x)g'(x), you are doing the reverse
>trig substitution
A tricky technique for sure. By substitution a trig formula in for x, you can use the properties of trig functions to manipulate the function into acceptable form.

Pic related is an example trig sub problem. Usually you will use this technique when you see something under a radical.

There are other techniques of course. For example, breaking sines and cosines into imaginary exponentials. But these are the main ones you should be familiar with

it's not that complicated

Basically factor x^2 - x + 1 using quadratic theorem. This will give you two numbers with imaginary parts
And as you know, any complex numbers can be expressed as an exponential.

Then using partial fractions you split the two factors apart into separate pieces

>Sup Veeky Forums does anyone have a good e book or some shit like that to learn how to find antiderivatives ?

When I was in Calc I and II I used UC Davis: math.ucdavis.edu/~kouba/ProblemsList.html

I love them because they have challenging problems with full step by step solutions. }

>what the fuck is the anti derivative of (x+1)/(x^2-x+1)

There are two ways to find it. One would be by partial fractions (look up this technique). Here this may not be too helpful because you have to factorize this polynomial and to do this you will need complex numbers.

A better way would be by realizing that:
[eqn] \frac{x+1}{x^2-x+1} = \frac{2x-1 + 3}{2(x^2-x+1)} = \frac{2x-1}{2(x^2-x+1)} + \frac{3}{2(x^2-x+1)} [/eqn]

For the fraction on the left first remember that in an integral you can take out that 2 in the denominator. Then note that the derivative of the denominator is the numerator so you get the known form [math] \frac{du}{u} [/math]. For the one on the left you have to do a little more work:

[eqn] \frac{1}{x^2-x+1} = \frac{1}{(x-\frac{1}{2})^2 + \frac{3}{4}} [/eqn]

First, if you are asking where 3 and the 2 went, remember that you can take out those constants when integrating.

Now, notice that what you have remaining there looks a lot like the derivative of the arctan function. You only need to do a little work with u-substitution to get there.

don't listen to him
This guy has his shit together

>Anti derivative
And here I've lived my whole life thinking it's called an integral.

Helping him is directly contributing to the board's overall decrease in health.

Pure mathematics student to the rescue:

Definition:
If the derivative of F(x) is g(x) then we say that F is an antiderivative of g

Theorem: The antiderivative of a function is not unique:
Proof: Let F be an antiderivative of g. Then F(x) + C is another antiderivative of g.

Definition:
The family of all antiderivatives of a function g is called the indefinite integral of g.

That is why when you find the indefinite integral, you are forced to put the + C. Because indefinite integral stands for a family of antiderivatives, not just one.

I prefer primatives

Kek

>does it even exist
There is an explicit algorithm for finding the anti derivatives of fractions of polynomials.

I suggest you google that.

Oh that's interesting, I didn't know that. I think the lines separating that are really blurred in Danish. They are just used interchangeably, even though here actually are different words now that I've looked it up.

That's normal for calc2
I just did AP Calc BC, skipped that bullshit. Fuck calc 2.

They are completely and totally interchangeable in practical contexts

It's normal if you aren't some autist

Here you are : the answers exists
But do not count on me for the proof