Rewrite
#4x^2-12x-7=(2x-3)^2-16#
#=4^2(((2x-3)/4)^2-1)#
Therefore,
#sqrt(4x^2-12x-7)=sqrt(4^2(((2x-3)/4)^2-1))#
#=4sqrt((((2x-3)/4)^2-1))#
The integral is
#I=int(dx)/(sqrt(4x^2-12x-7))=1/4int(dx)/(sqrt((((2x-3)/4)^2-1)))#
Let #u=(2x-3)/4#, #=>#, #du=dx/2#
Therefore,
#I=1/4int(2du)/(sqrt(u^2-1))=1/2int(du)/sqrt(u^2-1)#
Let #u=secv#, #=>#, #du=secvtanvdv#
#sec^2v-1=tan^2v#
The integral is
#I=1/2int(secvtanvdv)/sqrt(sec^2v-1)#
#=1/2intsecvdv#
#=1/2int(secv(secv+tanv)dv)/(secv+tanv)#
Let #w=secv+tanv#, #=>#, #dw=(secvtanv+sec^2v)dv#
Therefore,
#I=1/2int(dw)/w#
#=1/2ln(w)#
#=1/2ln(secv+tanv)#
#=1/2ln(u+sqrt(u^2-1))#
#=1/2ln((2x-3)/4+sqrt((((2x-3)/(4))^2)-1))#
#=1/2ln(|(2x-3)/4+sqrt((((2x-3)/(4))^2)-1)|)+C#
