Evaluate the following integrals

Did you miss the number in the second factor (x - 3)(x + ?)...?

Since you missed the factor in x^2/((x-3)(x+)) I'll call it d

x^2/((x-3)(x+d))=Ax/(x-3)+Bx/(x+d)
or
(Ax(x+d)+Bx(x-3))/((x-3)(x+d))=x^2/((x-3)(x+d))so the problem is to solve for A and B
and since the denominators are the same
Ax^2+Adx+Bx^2-3B=x^2
and you get two equations in two unknowns
A+B=1
and
Ad-3B=0
using substitution, elimination or whatever
A=3/(3+d) and B=d/(3+d)
so
x^2/((x-3)(x+d))=3/(3+d)x/(x-3)+d/(3+d)x/(x+d)
and you solve the integral by integrating x/(x-3) and x/(x+d) between 4 and 2 and multiplying these integrals by A and B respectively.

Rap