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).
