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

Note that x³+1 = (x+1)(x²-x+1).

You''ll wind up with a partial fraction decomposition with two logarithmic integrals, plus a trig integral.

Start as follows:
x/(x³+1) = A/(x+1) + (Bx+C)/(x²-x+1)

and find the values of A,B,C....

and solve.

I get 1/6(ln(x^2 -x +1) -1/3(ln(x+1)) + arctan(2x-1/root3)/root3

Please approve the answer.... :)

apply the formula
1+x^3=(1+x)(1+x^2-x)
and then apply partial fractions.
and then proceed.

x/(1+ x^3)
=1/x^2 / ( 1/x^3 + 1)
let 1/x^3 + 1 = t
then  -3/x^2 dx = dt
so the integrl reduces to
int( -1/3 * [dt/t]

=-1/3*ln|t| + c
=-1/3 * ln | 1/x^3 + 1 | + c