First note that
#int dx/(x+2)^(1/4)=int (x+2)^(-1/4)dx=4/3 (x+2)^(3/4)+C#.
Next, note that the "impropriety" in the definite integral occurs at #x=-2# because the function has a vertical asymptote there. Hence,
#int_{-2}^{14}dx/(x+2)^(1/4)=lim_{a->-2+}int_{a}^{14}dx/(x+2)^(1/4)#
assuming this limit exists, where the positive sign to the right of #-2# in the limit indicates that #a# is approaching #-2# from the right.
By the Fundamental Theorem of Calculus,
#int_{a}^{14}dx/(x+2)^(1/4)=4/3 (x+2)^(3/4)|_{a}^{14}#
#=4/3 * 16^(3/4) - 4/3(a+2)^(3/4)=32/3 - 4/3(a+2)^(3/4)#.
Since #lim_{a->-2+}(a+2)^(3/4)=0^(3/4)=0#, it follows that the improper integral converges to #32/3#.
We therefore write
#int_{-2}^{14}dx/(x+2)^(1/4)=32/3#.
