Just, the Integral of (x^3)/(x^3+1) dx
I know you can turn the denominator into (X+1)(X^2-x+1). What do you do from there? thanks =)
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Just, the Integral of (x^3)/(x^3+1) dx
I know you can turn the denominator into (X+1)(X^2-x+1). What do you do from there? thanks =)
I doublechecked this but I'm a bit rusty lately so perhaps someone else can verify it,
int dx/(x+1)(x^2-x+1)
You have an irreducible quadratic factor so the equation is
a/(x+1) + (bx+c)/(x^2-x+1) = 1
x=0 -> a+c=1
x=1 -> a+2b+2c=1
x=-1 -> 3a=1
Solving that system, a=1/3, b=-1/3, c=2/3
Now all you have to do is to solve the last part:
int 1/(3(x+1)) + ((-1/3)x + (2/3))/(x^2-x+1)
Hope this helps :)
dax918 is right for the part he did, but he left out an important detail. your integral is actually the integral of 1 minus all he said.
Note that
as dax918 said,
thus, your integral is
Thank you Jhevon for pointing that out, I forgot the fact that 3deltat already did the first step :)