As in the given algebraic fraction, we have the only denominator #1+x^2#, which cannot be factorized. As it is a quadratic function, in partial decomposition, the numerator would be in form #ax+b#. But it is already in this form. Hence we cannot convert them into further partial fractions.
Now let #u=1+x^2#, hence #du=2xdx#
Hence, #int(x-1)/(1+x^2)dx#
= #intx/(1+x^2)dx-int1/(1+x^2)dx#
Now #intx/(1+x^2)dx=int(du)/(2u)=1/2xxlnu=(ln(1+x^2))/2#
and #int1/(1+x^2)dx=tan^(-1)x#
(as if #tan^(-1)x=v#, #x=tanv# and
#(dx)/(dv)=sec^2v=1+tan^2v=1+x^2#, hence #(dx)/(1+x^2)=dv#
and #int(dx)/(1+x^2)=intdv=v=tan^(-1)x#)
Hence #int(x-1)/(1+x^2)dx=1/2ln(1+x^2)-tan^(-1)x#
