The function f(x)=x/|x|f(x)=x|x| equals x/x = 1xx=1 if x > 0x>0 and equals x/(-x)=-1x−x=−1 if x < 0x<0. It is undefined at 0. The "impropriety" for the integral int_{-5}^{3}x/|x|dx∫3−5x|x|dx therefore occurs at 00.
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.