Assuming you know how to do substitution :
Complete the square, you want to have #(ax+-b)^2+-1# form
#int 1/sqrt(3x^2-12x-5)dx#
#3x^2-12x-5#
#a = sqrt(3)x#
#b = 2sqrt(3)#
#a^2 = 3x^2#
#b^2=12#
#3x^2-12x+12-17#
#(sqrt(3)x-2sqrt(3))^2-17#
So we have
#int 1/sqrt(3x^2-12x-5)dx = int 1/sqrt((sqrt(3)x-2sqrt(3))^2-17)dx#
#= int 1/sqrt(17(1/17(sqrt(3)x-2sqrt(3))^2-1))dx#
#= 1/sqrt17int 1/sqrt(1/17(sqrt(3)x-2sqrt(3))^2-1)dx#
In fact, at this moment you can do #u = 1/sqrt17(sqrt(3)x-2sqrt(3)) #
#du = sqrt(3/17)dx#
#dx = sqrt(17/3)du#
#u^2=1/17(sqrt(3)x-2sqrt(3))^2#
#= 1/sqrt3int 1/sqrt(u^2-1)du#
You have the perfect derivate of #1/sqrt(3)arccosh(u)#
One form of the integral of #int 1/sqrt(3x^2-12x-5)dx#
is #1/sqrt(3)arccosh(1/sqrt(17)(sqrt(3)x-2sqrt(3)))#
But i don't like this form, so coming back to :
#1/sqrt17int 1/sqrt(1/17(sqrt(3)x-2sqrt(3))^2-1)dx#
#sqrt(3)x-2sqrt(3) = sqrt(17)/cos(u)#
note that #u = arccos(sqrt(17)/(sqrt(3)x-2sqrt(3)))#
#sqrt(3)dx = sqrt(17)tan(u)/cos(u)du#
#dx = sqrt(17/3)tan(u)/cos(u)#
You get
#1/sqrt3int 1/sqrt(1/(cos^2(u))-1)tan(u)/cos(u)dx#
#1/sqrt(1/(cos^2(u))-1) = 1/|tan(u)|#
#1/sqrt(3)int1/cos(u)du#
#1/sqrt(3)intcos(u)/cos^2(u)du#
let's #t = sin(u)#
#dt = cos(u)du#
#= 1/sqrt(3)int1/(1-t^2)dt#
do partial fraction to get
#1/sqrt(3)(int1/(2(t+1))dt-int1/(2(t-1))dt#)
which is
#1/(2sqrt(3))[ln(|t+1|)-ln(|t-1|)]#
Substitute back for #t = sin(u)#
#1/(2sqrt(3))[ln(|sin(u)+1|)-ln(|sin(u)-1|)]#
Substitute back for #u = arccos(sqrt(17)/(sqrt(3)x-2sqrt(3))) #
#1/(2sqrt(3))[ln(|sin(arccos(sqrt(17)/(sqrt(3)x-2sqrt(3))))+1|)-ln(|sin(arccos(sqrt(17)/(sqrt(3)x-2sqrt(3))))-1|)]#
assuming that
#sin(arccos(sqrt(17)/(sqrt(3)x-2sqrt(3)))) = sqrt(1-17/(sqrt(3) x-2 sqrt(3))^2)#
You finally have
#1/(2sqrt(3))[ln(|sqrt(1-17/(sqrt(3) x-2 sqrt(3))^2)+1|)-ln(|sqrt(1-17/(sqrt(3) x-2 sqrt(3))^2)-1|)]#
