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F(x) aln(x+1)+bln(x^2+1)+c*arc tgx
To find a,b andc.
2x/(1+x)(x^2+1) = A/(x+1) +B(x+1)(x^2+1)
2x/(x+1) = -1/(x+1)+(x+1)/(x^2+1)
2x/(x+1) = -1/(x+1) + (1/2)* 2x/(x^2+1) + 1/(x^2+1)
Now integrating, we get:
Int f(x) dx = Int 2xdx/(x+1)(x^2+1) = Int -1dx/(x+1) + Int(1/2)* 2xdx/(x^2+1) + Int dx/(x^2+1)
Int f(x) dx= -lnx +(1/2 ln (x^2+1) +arctan x
F(x) = (-1)lnx+ (1/2) ln(x^2+1) +1* arc tanx
Equating the coefficients , we get: a = -1 , b = 1/2 and c =1
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
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