#int x/sqrt(3x^2-6x+10) dx#
= #1/6int(6x-6)/sqrt(3x^2-6x+10)dx+6/sqrt3int(dx)/sqrt(x^2-2x+10/3)#
For the first part let #u=3x^2-6x+10# ten #du=(6x-6)dx#
and integral becomes #1/6int(du)/(sqrtu)=1/6*1/2*u^(1/2)#
= #1/12sqrt(3x^2-6x+10)#
Second part can be written as
#2sqrt3int1/sqrt((x-1)^2+7/3)dx=#
Now let #sqrt(3/7)(x-1)=tant# then #sqrt(3/7)dx=sec^2tdt# and our integral becomes
#2sqrt3sqrt(3/7)intsec^2t/sectdt=6/sqrt7ln|sect+tant|#
= #6/sqrt7ln|sqrt(3/7(x-1)^2+1)+sqrt(3/7)(x-1)|#
= #6/sqrt7ln|(1/7(3x^2-6x+10))+sqrt(3/7)(x-1)|#
and hence #int x/sqrt(3x^2-6x+10) dx#
= #1/12sqrt(3x^2-6x+10)+6/sqrt7ln|(1/7(3x^2-6x+10))+sqrt(3/7)(x-1)|#
