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