How do you integrate #(x-1)/(1+x^2)# using partial fractions?

1 Answer
Sep 8, 2016

#int(x-1)/(1+x^2)dx=1/2ln(1+x^2)-tan^(-1)x#

Explanation:

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#