#int_0^oo(x-1)/(x^3+x^2)\ dx#
Separate the integrand into partial fractions:
#(x-1)/(x^3+x^2)=A/x^2+B/x+C/(x+1)#
Now, multiply both sides by #x^2(x+1)#:
#x-1=A(x+1)+Bx(x+1)+Cx^2#
#\ \ \ \ \ " "=(B+C)x^2+(A+B)x+A#
Equating coefficients show that #{(B+C=0),(A+B=1),(A=-1):}#. Therefore, we have #A=-1,B=2,C=-2#.
So, #(x-1)/(x^3+x^2)=-1/x^2+2/x-2/(x+1)#
And our original problem becomes
#" "int_0^oo(x-1)/(x^3+x^2)\ dx#
#=int_0^oo-1/x^2+2/x-2/(x+1)\ dx#
#=-int_0^oodx/x^2+2int_0^oodx/x-2int_0^oodx/(x+1)#
#=[1/x+2ln|x|-2ln|x+1|]_0^oo#
#=[1/x+2ln|x/(x+1)|]_0^oo#
Evaluating at #x->oo# results to #0#, but it is undefined at #x=0#. Let's try evaluating as #x->0^+#.
#=0-lim_(x->0^+)(1/x+2ln|x/(x+1)|)#
#=-lim_(x->0^+)(1/x+2ln|x|-2ln|(x+1)|)#
#=-lim_(x->0^+)(1/x+2ln|x|)#
#=-lim_(x->0^+)((1+2xln|x|)/x)#
#=-lim_(x->0^+)((1+2ln|x^x|)/x)#
The limit is of the form #1/0^+#, so the final result is
#-oo#
We can check it with the graph of the function
graph{(x-1)/(x^3+x^2) [-0.5,10,-5,0.5]}
