Show that `int_a^b(|x|)/x dx=|b|=|a|dot`

Explanation

Case I: If `0lealtb`, then `|x|//x=1`
`:.I=int_(a)^(b)1dx=b-a=|b|-|a|`
Case II: If `altble0`, then `|x|=-x`
`:. I=int_(a)^(b)(-x)/xdx=int_(a)^(b)(-1)dx`
`=[-x]_(a)^(b)=b-(-a)=|b|-|a|`
Case III: If `alt0ltb`
then `|x|=-x` when `altxlt0`
and `|x|=x`, when `altxltb`
`:.I=int_(a)^(|x|)/xdx=int_(a)^(0)(|x|)/x dx+int_(0)^(b)(|x|)/x dx`
`=int_(a)^(0)(-x)/x dx+int_(0)^(b) x/x dx`
`=int_(a)^(0)(-1)dx+int_(0)^(b) 1 dx`
`=[-x]_(a)^(0)+[x]_(0)^(b)=a+b=b-(-a)=|b|-|a|`
Hence, in all the cases `I=int_(a)^(b)(|x|)/x dx=|b|-|a|`.