We want to find #int_-4^3(x+absx)/2tan^-1(3x)dx#
We start by splitting the integral where #x# is always negative and #x# is always positive
#int_-4^3(x+absx)/2tan^-1(3x)dx=#
#int_-4^0(x+absx)/2tan^-1(3x)dx+int_0^3(x+absx)/2tan^-1(3x)dx#
Now, the definition of #absx# is
#absx={{:(x, "if " x>=0),(-x, "if " x<0):}#
So our integral simplifies to
#int_-4^0(x-x)/2tan^-1(3x)dx+int_0^3(x+x)/2tan^-1(3x)dx#
#=int_0^3xtan^-1(3x)dx#
To make it a little bit easier, we substitute #u=3x# and #du=3dx# to get
#1/9int_0^9utan^-1(u)du#
We now use integration by parts to evaluate the integral
Let #v=tan^-1(u)rArrdv=1/(1+u^2)du#
and #dw=urArrw=1/2u^2#
So
#1/9int_0^9utan^-1(u)du=1/9([1/2u^2tan^-1(u)]_0^9-1/2int_0^9u^2/(1+u^2)du)=1/9(81/2tan^-1(9)-1/2int_0^9 1-1/(1+u^2)du)=1/9(81/2tan^-1(9)-1/2[u-tan^-1(u)]_0^9)=1/9(1/2(81tan^-1(9)-9+tan^-1(9)))=#
#-1/2+41/9tan^-1(9)#