First a few comments:
If we say #F'(x) = 1/(1+x^3)#,
we want to find F(x) where #intF'(x)dx=F(x)+C#.
To simplify I am skipping the constant C.
I will suppose you have a list of standard integration expressions, like in https://en.wikipedia.org/wiki/Lists_of_integrals
Therefore, I will suppose you know that
#int1/xdx = ln|x|#
and
#int1/(x^2+1)dx = arctan x#
I will, therefore, take these as given.
#int1/(x^3+1)dx# (plase the unknown first)
=#int1/[(x+1)(x^2-x+1)]dx# (factorise the denominator)
We observe that #(x+1)(x-2) = (x^2-x-2) = (x^2-x+1)-3#
So #(x^2-x+1) - (x+1)(x-2) = 3#
We, therefore, can write our expression as:
#int3/[3(x+1)(x^2-x+1)]dx = int[(x^2-x+1) - (x+1)(x-2)]/[3(x+1)(x^2-x+1)]# which we can split in two:
(1) #=1/3int1/(x+1)dx - 1/3int(x-2)/(x^2-x+1)dx#
In the first part in (1) we see:
#int1/(x+1)dx =int1/udu# (where u=x+1)
(2) =#ln(u) = ln(x+1)#
In the second part in (1) we have:
If #u = x^2-x+1#
we get #u'=2x-1#
i.e. #dx=1/(2x-1)du#
In addition we have #x-2 = 1/2(2x-1)-3/2#
Using this we get in (1):
#int(x-2)/(x^2-x+1)dx#
(3) =#1/2int(2x-1)/(x^2-x+1)dx-3/2int(1/(x^2-x+1))dx#
Substitute #u = x^2-x+1#, #dx=1/(2x-1)du# in the first part:
#int(2x-1)/(x^2-x+1)dx#
(4) =#int1/udu=ln(u) = ln(x^2-x+1)#
The second part in (3):
#int1/(x^2-x+1)dx#
The numerator can be written:
#(x-1/2)^2+3/4#
Therefore
#int1/(x^2-x+1)dx=int1/((x-1/2)^2+3/4)dx#
If we substitue #u=(2x-1)/sqrt(3)#, #dx=sqrt(3)/2du# in this, we get
#=int(2sqrt(3))/3u^2+3 = 2/sqrt(3)int1/(u^2+1)du#
#=2/sqrt(3)arctan(u)#
#=2/sqrt(3)arctan((2x-1)/sqrt(3))#
Plug this into (3):
(3) =#1/2int(2x-1)/(x^2-x+1)dx-3/2int(1/(x^2-x+1))dx#
#=1/2ln(x^2-x+1) - sqrt(3)arctan((2x-1)/sqrt(3))#
Plug this into (1):
#=1/3int1/(x+1)dx - 1/3int(x-2)/(x^2-x+1)dx#
#=-ln(x^2-x+1)/6+arctan((2x-1)/sqrt(3))+ln(x+1)/3#
Last step: apply the absolute value to the arguments of the logarithm functions:
#=-(ln|x^2-x+1|)/6+arctan((2x-1)/sqrt(3))+ln|x+1|/3#
