The function #f(x)=x/|x|# equals #x/x = 1# if #x > 0# and equals #x/(-x)=-1# if #x < 0#. It is undefined at 0. The "impropriety" for the integral #int_{-5}^{3}x/|x|dx# therefore occurs at #0#.
Assuming the integrals exist (converge), we can write #int_{-5}^{3}x/|x|dx=lim_{b->0-}int_{-5}^{b}x/|x|dx+lim_{a->0+}int_{a}^{3}x/|x|dx# (the notation means #b# approaches 0 "from the left" and #a# approaches 0 "from the right").
Now
#lim_{b->0-}int_{-5}^{b}x/|x|dx=lim_{b->0-}int_{-5}^{b}(-1)dx=lim_{b->0-}(-x)|_{-5}^{b}#
#=lim_{b->0-}(-b-(-(-5)))=lim_{b->0-}(-b-5)=-5#
and
#lim_{a->0+}int_{a}^{3}x/|x|dx=lim_{a->0+}int_{a}^{3}1dx=lim_{a->0+}(x)|_{a}^{3}#
#=lim_{a->0+}(3-a)=3#.
Therefore, these integrals converge and
#int_{-5}^{3}x/|x|dx=-5+3=-2#.
In the end, you should know that a finite number of "jump discontinuities " over a finite interval will not cause an improper integral over that interval to diverge.
