F(x) = aln(x+1) + bln(x^2+1) + carctgx f(x) = 2x/(x+1)(x^2+1) Find a, b, c so f to be first derivative of F

F'(X)=f(x)=[a ln(x+1)]' + [b ln(x^2+1)]' + [c
arctg(x)]'

F'(X)= a/(x+1) + 2bx/(x^2+1) + c/(x^2 +
1)

We'll try to write f(x) as a sum of simple
fraction:

2x/(x+1)(x^2+1)=A/(x+1) +
(Bx+C)/(x^2+1)

In order to have the same denominator in the
right side of the equal we have to multiply A with (x^2+1) and (Bx+C) with
(x+1).

2x=Ax^2 + A + Bx^2 + Cx +
Bx+C

2x= x^2(A+B) + x(C+B) +
A+C

Two expressions are identical, if the correspondent
terms from the both sides of equal are similar.

(A+B)=0,
A=-B, B=1

(C+B)=2, C-A=2,C+C=2,
C=1

A+C=0, A=-C,
A=-1

2x/(x+1)(x^2+1)=-1/(x+1) +
x+1/(x^2+1)

a/(x+1) + 2bx/(x^2+1) + c/(x^2 +
1)=2x/(x+1)(x^2+1)

a/(x+1) + 2bx/(x^2+1) + c/(x^2 +
1)=-1/(x+1) + x+1/(x^2+1)

a=-1, b=1,
c=1