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)x−2(x+√3)(x−√3)(x−3)(x−1)=Ax+√3+Bx−√3+Cx−3+Dx−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)DA(x−√3)(x−3)(x−1)+(x+√3)B(x−3)(x−1)+(x+√3)(x−√3)C(x−1)+(x+√3)(x−√3)(x−3)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)A(x3−(4+√3)x2+(3+4√3)x−3√3)+B(x3−(4−√3)x2+(3−4√3)x+3√3)+C(x3−x2−3x+3)+D(x3−3x2−3x+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.
