determine a, b, c if F'=f f=2x/(x+1)(x^2+1) F=a ln (1+x) + b ln ( 1+x^2) + c arctanx

We know from enunciation
that:

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 write f(x) as an addition of simple
quotients:

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

In order to have the same denominator to the
right side, we'll  multiply A by (x^2+1) and (Bx+C) by
(x+1).

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

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

To find A,B,C, we'll set the constraint that two
expressions are identical, if the correspondent terms from the both sides are
equal.

(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

F(x)= -ln (1+x) + ln( 1+x^2) + arctan
x

F(x)= ln [1/(1+x)] + ln ( 1+x^2) +
arctanx

F(x)= ln
[(1+x^2)/(1+x)] + arctanx