First we need to decompose this function to a sum of partial fractions (we assume it can be done):
#(x-2)/((x+sqrt3)(x-sqrt3)(x-3)(x-1))=color(red)A/(x+sqrt3)+color(green)B/(x-sqrt3)+color(blue)C/(x-3)+color(orange)D/(x-1)#
but after bringing it to common denominator, the numerator equals
#color(red)A(x-sqrt3)(x-3)(x-1)+(x+sqrt3)color(green)B(x-3)(x-1)+(x+sqrt3)(x-sqrt3)color(blue)C(x-1)+(x+sqrt3)(x-sqrt3)(x-3)color(orange)D#
Using Vieta's formulas for 3rd degree polynomials, we get
#color(red)A(x^3-(4+sqrt3)x^2+(3+4sqrt3)x-3sqrt3)
+color(green)B(x^3-(4-sqrt3)x^2+(3-4sqrt3)x+3sqrt3)
+color(blue)C(x^3-x^2-3x+3)+color(orange)D(x^3-3x^2-3x+9)#
Combining like terms:
#(color(red)A+color(green)B+color(blue)C+color(orange)D)x^3#
#-((4+sqrt3)color(red)A+(4-sqrt3)color(green)B+color(blue)C +3color(orange)D)x^2
+((3+4sqrt3)color(red)A+(3-4sqrt3)color(green)B-3color(blue)C-3color(orange)D)x
+(-3sqrt3color(red)A+3sqrt3color(green)B+3color(blue)C
+9color(orange)D)#
but we remember it's equal to #x-2#, so we have
#{(color(red)A+color(green)B+color(blue)C+color(orange)D=0),
((4+sqrt3)color(red)A+(4-sqrt3)color(green)B+color(blue)C +3color(orange)D=0),
((3+4sqrt3)color(red)A+(3-4sqrt3)color(green)B-3color(blue)C-3color(orange)D=1),
(-3sqrt3color(red)A+3sqrt3color(green)B+3color(blue)C
+9color(orange)D=-2):}#
Solving the system gives:
#color(red)A=1/12, color(green)B=1/12, color(blue)C=1/12, color(orange)D=-1/4#
So we have:
#f(x)=(1/12)/(x+sqrt3)+(1/12)/(x-sqrt3)+(1/12)/(x-3)-(1/4)/(x-1)#
Integral of sum is sum of integrals, so here's integral of the first term:
#int(1/12)/(x+sqrt3)dx=1/12int1/(x+sqrt3)dx#
substituting #u=x+sqrt3# and #du=dx# we have
#1/12int1/udu=1/12lnu+C=1/12ln(x+sqrt3)+C#
The other 3 terms are done similarly.
