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
