Question

integral x^3/(x^3 +1)

Answer 1

This is a bit ugly. Remember that x^3 + 1 = (x+1)(x^2 - x + 1)

This points you down the partial fractions route, and that's where you need to go.

Answer 2

i think you can do this by the substitution u = x^3

Answer 3

so far i've gotten it to:

x -1/3ln{x+1} + integral of (-x/3 + 2/3)/(x^2 -x +1), is that right? and if so how do i do i integrate it from there

Answer 4

You can first simplify it to \(1-\frac{1}{x^3+1}=1-\frac{1}{(x+1)(x^2-x+1)}\). Then you should use partial fractions to expand the second term.

Answer 5

So x^3/(x^3 + 1) = 1 -1/(x^3+1)
so the integrating the RHS, there is definitely an x.

Now we need to write down the partial fraction expansion of the 1/(x^3+1), which is
\[\frac{2-x}{3(x^2 - x +1)} + \frac{1}{3(x+1)}\]

So yes, I agree with you so far.

Now you separate the two terms over x^2 - x + 1. One will give you a log, the other an arctan.

Answer 6

just be careful with the signs: I might have made an error, or you've made an error as there's a small inconsistency between us.

Answer 7

how would i do the integral of (-x/3 + 2/3)/(x^2 -x +1

Answer 8

the integral of x/(x^2 - x + 1) = x/((x-1/2)^2 + (sqrt(3)/2)^2)

Now you make a substitution of u = x - 1/2

The integral of 1/((x^2 - x + 1) = 1/((x-1/2)^2 + (sqrt(3)/2)^2)

is an arctan: set x = 1/2 + sqrt(3)/2.tan u

Answer 9

but what happened to the -1/3 and 2/3?

Answer 10

I took them out just to simply the description, but obviously you'll need to put them back in an scale back up the integrals.

Answer 11

...just to simplify...

Answer 12

could i just multiply the equation by 3 to get rid of the rations?

Answer 13

Sure ... whatever makes it easier for you. Just remember to scale it back at the end.

Answer 14

i don't see how you broke up the denominator though?

Answer 15

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