Fx alnx1 blnx21 carctgx fx 2xx1x21 Find a b c so f to be

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

Solution

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