# 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

hala718 | High School Teacher | (Level 1) Educator Emeritus

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

F = aln (1+x) + b ln(1+x^2) +c*arctanx

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

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

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

= [(a+2b)x^2 + (2b+1)x + (a+c)]/(1+x^2)(1+x)

But F' = f

==> (a+2b)x^2 + (2b+1)x + (a+c) = 2x

==> a+2b = 0......(1)

==> 2b+1 = 2

==> b = 1/2

==> a = -2b = -2*1/2 = -1

==> a = -1

==> a+c = 0.........(3)

==> c = 1

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f  = 2x/(x+1)(x^2+1)

Solution:

Let us split the right side into partial fractions.

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

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

To determine A,B,C

Put x= 0: 0= A +C Or A = -C.

Put x =-1 in (1): 2 = 2A  Or A = 1. So  C = -A = -1.

Put   x=1 , 2 = A(1+1)+(B*1-1)(1+1) = 2 +(B-1)2 = 2B. Or B= 1.

Therefore

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

fF(x)=   Integral [1/(x+1)]dx  + Integral {dlog(x^2)x/(x^2+1)  -1/(1+x^2)}dx

= log(1+x) + (1/2) d[log (x^2+1)]/(x^2+1) - 1/(1+x^2)

= log(1+x)+(1/2) -arc tanx  which is given to be equal to aln(x+1)+bln(x^2+1) +c rrc tanx.

So by equating coefficients of similar expressions, we get:

a = 1, b = 1/2 and c = -1.