Begin by factorizing the denominator:
#1+x^3=(x+1)(x^2-x+1)#
Now we can do partial fractions:
#1/(1+x^3)=1/((x+1)(x^2-x+1))=A/(x+1)+(Bx+C)/(x^2-x+1)#
We can find #A# using the cover-up method :
#A=1/((text(////))((-1)^2+1+1))=1/3#
Next we can multiply both sides by the LHS denominator:
#1=1/3(x^2-x+1)+(Bx+C)(x+1)#
#1=1/3x^2-1/3x+1/3+Bx^2+Bx+Cx+C#
#1=(1/3+B)x^2+(B+C-1/3)x+(C+1/3)#
This gives the following equations:
#1/3+B=0 ->B=-1/3#
#C+1/3=1->C=2/3#
This means that we can rewrite our original integral:
#int\ 1/(1+x^3)\ dx=1/3int\ 1/(x+1)-(x-2)/(x^2-x+1)\ dx#
The first integral can be done using an explicit u-substitution, but it's rather clear that the answer is #ln|x+1|#:
#1/3(ln|x+1|-int\ (x-2)/(x^2-x+1)\ dx)#
We can split the remaining integral into two:
#int\ (x-2)/(x^2-x+1)\ dx=1/2int\ (2x-4)/(x^2-x+1)\ dx=#
#=1/2(int\ (2x-1)/(x^2-x+1)\ dx-int\ 3/(x^2-x+1)\ dx)#
The reason for the trickery with the multiplying and dividing by #2# is to make the left hand denominator easier to use u-substitution on.
I'll call the left integral Integral 1 and the right integral Integral 2
Integral 1
#int\ (2x-1)/(x^2-x+1)\ dx#
Since we already prepared this integral for substitution, all we need to do is substitute #u=x^2-x+1#, and the derivative is #2x-1#, so we divide by that to integrate with respect to #u#:
#int\ cancel(2x-1)/(cancel(2x-1)*u)\ du=int\ 1/u\ du=ln|u|+C=ln|x^2-x+1|+C#
Integral 2
#3int\ 1/(x^2-x+1)\ dx#
We want to get this integral into the form:
#int\ 1/(1+t^2)\ dt=tan^-1(t)+C#
To do this, we need to complete the square for the denominator:
#x^2-x+1=(x-1/2)^2+k#
#x^2-x+1=x^2-x+1/4+k#
#k=3/4#
#3int\ 1/(x^2-x+1)\ dx=3int\ 1/((x-1/2)^2+3/4)\ dx#
We want to introduce a u-substitution such that:
#(x-1/2)^2=3/4u^2#
#x-1/2=sqrt3/2u#
#x=sqrt3/2u+1/2#
We multiply by the derivative with respect to #u# to integrate with respect to #u#:
#dx/(du)=sqrt(3)/2#
#3*sqrt3/2int\ 1/(3/4u^2+3/4)\ du=3sqrt3/2*1/(3/4)int\ 1/(u^2+1)\ du=#
#=2sqrt3tan^-1(u)+C=2sqrt3tan^-1((2x-1)/sqrt3)+C#
Completing the original integral
Now that we know the answer to Integral 1 and Integral 2, we can plug them back into the original expression to get our final answer:
#1/3(ln|x+1|-1/2ln|x^2-x+1|+sqrt3tan^-1((2x-1)/sqrt3))+C=#
#=1/3ln|x+1|-1/6ln|x^2-x+1|+sqrt3/3tan^-1((2x-1)/sqrt3)+C#