Sup /adv/ any math fags over here ? I need some advice on how

>antiderivative

We don't want to do your high school homework for you. That being said, are you sure that's correct? The denominator has no rational roots.

>>18229356
Good job making fun of him for an integral that clearly isn't a high school level problem haha
>>18229306
You gotta split that shit man:
1.) int(x/(x^2-x+1))+int(1/(x^2-x+1))

2.) Then, you can integrate the left side as a typical log (1/2*log(x^2-x+1)+1/2)

3.) The right side is a bit more tricky, but I think you can do it as a trig substitution for x=tan(u), dx=sec^2(u) du , but I'm still working on that side

>>18229306
https://www.wolframalpha.com

>>18229367
Nevermind, I messed up a little - you do need to split it, but there needs to be a step before it:

1.) Write out x+1 as 1/2*(2x+1) + 3/2
This step is done because you need the '2x+1' term since it is the derivative of the denominator. This gives you:

int((2x+1)dx/(2(x^2-x+1)))+int(3dx/(2(x^2-x+1)))

2.) integrate the left side to get 1/2ln(x^2-x+1)

3.) For 3/2 int(dx/(x^2-x+1)), you need to complete the square:
ax^2+bx+c ---> a(x+b/2a)^2+c-b^2/2a
where a=1, b=-1, and c=1
which gives (x-1/2)^2-3/4
in full: 3/2 int(dx/((x-1/2)^2-3/4)))

4.) We want to get this into a 1/(1+u^2) form so that we can do the arctan(x) integral, so we need to make a sub that makes it look like that.

>>18229367
I did way more complex integrals in high school, is this what Americans really think?

>>18229385
I love wolfram alpha

>>18229306
Try partial fractions or maybe some trig sub. If nothing else integratuon by parts. I lost my math powers so my bad OP.

X^2 - X + 1 = (X - 1/2)^2 + 3/4
Now use substitute 3/4sinh u ^ 2 = (X - 1/2)^2. This should give you a cosh^2 for the denominator, and should now be a lot easier:
Root(3)/2 sinh u = X - 1/2
Root(3)/2 cosh u du = dx
So we have
Int((X+1)/(X^2-X+1) dx) = int(something / 3/4cosh u ^2)

Where something is Root(3)/2 cosh u * Root(3)/2 sinh u + 3/2
This should give an integral in terms of tanh and sech, which are both integrable.

I think you should get int(tanh u + root(3)sech u)
= ln(cosh u) + root(3)arctan(sinh u)
This is assuming you know your hyperbolic integrals. They are immensely useful.