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#
