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!
