Whenever we have a quadratic in the denominator and no #x#'s in the numerator, we want to get the integral into the following form:
#int\ 1/(1+t^2)\ dt=tan^-1(t)+C#
In our case, we can do this by completing the square and then using a substitution.
#x^2+x+1=(x+1/2)^2+k#
#x^2+x+1=x^2+x+1/4+k#
#k=3/4#
#x^2+x+1=(x+1/2)^2+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#
We can solve for #x# to figure out what this substitution needs to be:
#x+1/2=sqrt3/2u#
#x=sqrt3/2u-1/2#
To integrate with respect to #u#, we multiply by the derivative of #x# with respect to #u#:
#dx/(du)=sqrt3/2#
#3int\ 1/((x+1/2)^2+3/4)\ dx=3*sqrt3/2int\ 1/(3/4u^2+3/4)\ du=#
#=3*sqrt3/2int\ 1/(3/4(u^2+1))\ du=3*sqrt3/2*4/3int\ 1/(u^2+1)\ du=#
#=2sqrt3tan^-1(u)+C#
We can now solve for #u# in terms of #x# to resubstitute:
#u=(2x+1)/sqrt3#
This means our final answer is:
#2sqrt3tan^-1((2x+1)/sqrt3)+C#
