4/(x^2+1)-(2x/(x^2+1)^2)=0

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Solution for 4/(x^2+1)-(2x/(x^2+1)^2)=0 equation: 


D( x )

x^2+1 = 0

(x^2+1)^2 = 0

x^2+1 = 0

x^2+1 = 0

1*x^2 = -1 // : 1

x^2 = -1

(x^2+1)^2 = 0

(x^2+1)^2 = 0

1*x^2 = -1 // : 1

x^2 = -1

x in (-oo:+oo)

4/(x^2+1)-((2*x)/((x^2+1)^2)) = 0

4/(x^2+1)-2*x*(x^2+1)^-2 = 0

4/(x^2+1)+(-2*x)/((x^2+1)^2) = 0

(4*(x^2+1))/((x^2+1)^2)+(-2*x)/((x^2+1)^2) = 0

4*(x^2+1)-2*x = 0

4*x^2-2*x+4 = 0

4*x^2-2*x+4 = 0

2*(2*x^2-x+2) = 0

2*x^2-x+2 = 0

DELTA = (-1)^2-(2*2*4)

DELTA = -15

DELTA < 0

2 = 0

2/((x^2+1)^2) = 0

2/((x^2+1)^2) = 0 // * (x^2+1)^2

2 = 0

x belongs to the empty set

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