Split the rational function:
#x/sqrt(3x^2-6x+17) = ((1/6(6x-6))+1)/sqrt(3x^2-6x+17) #
so that using the linearity of the integral:
#int x/sqrt(3x^2-6x+17)dx = 1/6 int (6x-6)/sqrt(3x^2-6x+17)dx + int dx/sqrt(3x^2-6x+17) #
Solve now the first integral by substituting:
#u = 3x^2-6x+17#
#du = (6x-6)dx#
so that:
# 1/6 int (6x-6)/sqrt(3x^2-6x+17)dx = 1/3 int (du)/(2sqrtu) = 1/3 sqrtu+C#
and undoing the substitution:
# 1/6 int (6x-6)/sqrt(3x^2-6x+17)dx = 1/3 sqrt(3x^2-6x+17)+C#
Solve the second integral by complete the square at the denominator:
#int dx/sqrt(3x^2-6x+17) = int dx/sqrt(3(x^2-2x+1)+14)#
#int dx/sqrt(3x^2-6x+17) = int dx/sqrt(3 (x-1)^2+14)#
#int dx/sqrt(3x^2-6x+17) = 1/sqrt14 int dx/sqrt(( (sqrt3 (x-1))/sqrt14)^2+1)#
Now substitute:
#v = (sqrt3 (x-1))/sqrt14#
#dv = sqrt3/sqrt14dx#
#int dx/sqrt(3x^2-6x+17) = 1/sqrt3 int (dv)/sqrt(v^2+1)#
and then:
#v = tant# with #t in (pi/2,pi/2)#
#dv = sec^2t dt#
#int dx/sqrt(3x^2-6x+17) = 1/sqrt3 int (sec^2tdt)/sqrt(tan^2t+1)#
Using the trigonometric identity:
#tan^2t+1 = sec^2t#
and as #sect > 0 # for #t in (pi/2,pi/2)#:
#sqrt(tan^2t+1) = sect#
so:
#int dx/sqrt(3x^2-6x+17) = 1/sqrt3 int sectdt#
#int dx/sqrt(3x^2-6x+17) = 1/sqrt3 ln abs (sect+tant) +C#
#int dx/sqrt(3x^2-6x+17) = 1/sqrt3 ln abs (v+sqrt(v^2+1)) +C#
#int dx/sqrt(3x^2-6x+17) = 1/sqrt3 ln abs ((sqrt3 (x-1))/sqrt14+sqrt(((sqrt3 (x-1))/sqrt14)^2+1)) +C#
#int dx/sqrt(3x^2-6x+17) = 1/sqrt3 ln (sqrt3 abs(x-1)+sqrt(3x^2-6x+17) ) +C#
Putting the results together:
#int x/sqrt(3x^2-6x+17)dx = 1/3 sqrt(3x^2-6x+17) + 1/sqrt3 ln (sqrt3 abs(x-1)+sqrt(3x^2-6x+17) ) +C#
