f(x) has this decomposition in partial fractions

     =  + .    (1)


with some real coefficients "a", "b" and "c".  

These coefficients are unknown now, and the rest part of this post is to determine them.



Multiply both sides of equality (1) by  . You will get

    6 + 7x = (ax+b)*(2-x) + .    (2)


At x= 2,  equation  (2)  gives

    6 + 7*2 = c*(1+2^2),  or  20 = 5c,  or  c= 20/5 = 4.


At x= 0 and c= 4,  equation  (2)  gives

    6 + 7*0 = (a*0+b)*(2-0) + 4*(1+0^2),  or  6 = 2b + 4;  then  6-4 = 2b,  2 = 2b,  b = 2/2 = 1.


At x= 1 and c= 4, b= 1,  equation  (2)  gives

    6 + 7*1 = (a*1+1)*(2-1) + 4*(1+1^2),  or  13 = (a+1) + 4*2;  then  13 - 8 = a+1;  5 = a+1;  a= 5-1 = 4.


ANSWER.   =  + .