express x×x+x+4/(1-x)(x×x+1) into partial fraction

Answer 1

Assuming parentheses around the denominator have been omitted,
the expression is
x²+x+4/[(1-x)(x²+1)]
partial fractions for the third term is required, namely
4/[(1-x)(x²+1)]
=-4/[(x-1)(x²+1)]
=A/(x-1)+(Bx+C)/(x²+1)

By adding the right-hand side, we have
A(x²+1)+(Bx+C)(x-1)=-4

By matching coefficients of x²,x and constant term, we obtain
A+B=0
B+C=0
A-C=-4
Solving, A=-2, B=C=2
thus the partial fraction is
(2x+2)/(x²+1)-2/(x+1)

Be sure to add the partial fractions to the first two terms x²+x.

Answer 2

And assuming the usual sloppiness in the numerator, we have

(x²+x+4)/[(1-x)(x²+1)]
= (2x+1)/(x²+1) + 3/(1-x)

amazing how little difference it made, eh?