Complete the square at the denominator:
#x^2-4x+3 = (x-2)^2 -1#
and subtitute:
#x-2 = sect#
#dx = tant sect dt#
Note also that:
#x^2-4x+3 = (x-3)(x-1)#
and the integrand function is defined for
#x^2-4x+3 >0#
that is for #x in (-oo,1) uu (3,+oo)#. We will start by considering the interval #x in (3,oo)# that corresponds to #t in (0,pi/2)#.
Then:
#int dx/sqrt(x^2-4x+3) = int dx/sqrt((x-2)^2 -1)#
#int dx/sqrt(x^2-4x+3) = int (sect tant dt)/sqrt(sec^2t -1)#
Use now the trigonometric identity:
#sec^2t -1 = tan^2t#
and as for #t in (0,pi/2)# the tangent is positive:
#sqrt(sec^2-1) = tant#
so:
#int dx/sqrt(x^2-4x+3) = int (sect tant dt)/tant#
#int dx/sqrt(x^2-4x+3) = int sectdt#
The integral of #sect# should be known, but here is how is calculated:
#int sectdt = int sect ((sect+tant)/(sect+tant))dt#
#int sectdt = int ((sec^2t+tantsect)/(sect+tant))dt#
#int sectdt = int (d(tant+sect))/(sect+tant)dt#
#int sectdt = ln abs(sect+tant) +C#
Then:
#int dx/sqrt(x^2-4x+3) = ln abs(sect+tant) +C#
and to undo the substitution we note that:
#sect = (x-2)#
#tant = sqrt(sec^2t-1) = sqrt((x-2)^2-1) = sqrt(x^2-4x+3)#
so:
#int dx/sqrt(x^2-4x+3) = ln abs(x-2+sqrt(x^2-4x+3)) +C#
and by direct verification we can check that the solution is valid also for #x in (-oo,1)#.
