As #x/(1+absx)# is an odd function integrated over an integral symmetric with respect to #x=0# we can immediately deduce that:
#int_(-3)^3 (xdx)/(1+absx) = 0#
In fact using the properties of integrals:
#int_(-3)^3 (xdx)/(1+absx) = int_(-3)^0 (xdx)/(1+absx) + int_0^3 (xdx)/(1+absx) #
Now for #x in (-3,0)# we have #abs x = -x# while for #x in (0,3)# we have #absx = x#, so:
#int_(-3)^3 (xdx)/(1+absx) = int_(-3)^0 (xdx)/(1-x) + int_0^3 (xdx)/(1+x) #
Substitute in the first integral #t=-x#:
# int_(-3)^0 (xdx)/(1-x) = int_(3)^0 (tdt)/(1+t) = -int_0^3 (tdt)/(1+t) #
So:
#int_(-3)^3 (xdx)/(1+absx) = -int_0^3 (tdt)/(1+t) + int_0^3 (xdx)/(1+x) =0#
We can also find a primitive of the function: for #x > 0#:
#int (xdx)/(1+absx) = int (xdx)/(1+x) #
#int (xdx)/(1+absx) = int (1+x-1)/(1+x) dx#
#int (xdx)/(1+absx) = int dx - int 1/(1+x) dx#
#int (xdx)/(1+absx) = x - ln abs (1+x)+C #
while for #x<0#:
#int (xdx)/(1+absx) = int (xdx)/(1-x) #
#int (xdx)/(1+absx) = int (x-1+1)/(1-x) dx#
#int (xdx)/(1+absx) = - int dx + int 1/(1-x) dx#
#int (xdx)/(1+absx) = -x - ln abs (1-x)+C #
So in general:
#int (xdx)/(1+absx) = absx - ln (1+absx)+C #
