We split the Integrand as :
#(3x^2-1)/((x^2+2)(x-3))=A/(x-3)+(Bx+C)/(x^2+2),......................(star)#
where, #A,B,C in RR#.
#A# can be quickly determined by Heavyside"s Cover-up Method as
#A=[(3x^2-1)/(x^2+2)]_(x=3)=26/11#
Simplifying #(star)# & comparing rational polys. on both sides, we get,
#A(x^2+2)+(Bx+C)(x-3)=3x^2-1.....................(1)#
Taking #x=0, &, A=26/11# in #(1)#,
#26/11(2)+C(-3)=-1rArrC=21/11#.
#(1), x=1, A=26/11, C=21/11#
#rArr26/11(3)+(B+21/11)(-2)=2rArrB=7/11#.
With these #A,B,C#, we have, #int(3x^2-1)/((x^2+2)(x-3))dx#,
#=26/11int1/(x-3)dx+1/11int(7x+21)/(x^2+2)dx#,
#=26/11ln|x-3|+7/11*1/2int(2x)/(x^2+2)dx+21/11int1/(x^2+2)dx#,
#=26/11ln|x-3|+7/22int(d(x^2+2))/(x^2+2)+21/11*1/sqrt2*tan^(-1)(x/sqrt2)#,
#=26/11ln|x-3|+7/22ln(x^2+2)+21/(11sqrt2)tan^(-1)(x/sqrt2)+K#.
Enjoy Maths!