The problem is up the upper bound, as at x=3, the integrand has a discontinuity (vertical asymptote) due to division by 0.
Therefore, we want to replace the upper bound with t and take the limit of the integral as t->3^-. The reason we're taking the left-hand limit is because we're working with the upper bound, which we can only ever approach from the left in our interval of integration.
Lim_(t->3^-)int_2^tdx/(x-3)^(1/3)
There's not much to evaluating the actual integral. intdx/(x-3)^(1/3) could be rewritten as int(du)/u^(1/3)=intu^(-1/3)du=3/2u^(2/3)=3/2(x-3)^(2/3), where u=x-3, du=dx. We're leaving out the constant of integration because we're going to be taking a definite integral;
Lim_(t->3^-)3/2(x-3)^(2/3) where we must first evaluate 3/2(x-3)^(2/3) from 2 to t, which gives us:
3/2(t-3)^(2/3)-3/2(2-3)^(2/3)=3/2(t-3)^(2/3)-3/2
Because 3/2(-1)^(2/3)=3/2root(3)(-1^2)=3/2sqrt(1)=3/2
Lim_(t->3^-)3/2(t-3)^(2/3)-3/2=3/2(0^(2/3))-3/2=0-3/2=-3/2