Given the result of addition 1/(x^2+x) what are the two elementary fractions?
We have 1/(x^2+x). Now this has to be expressed as the sum of two fractions
=> 1/ (x + 1)*x
Let this be equal to
A / x + B / (x +1)
=> [A(x + 1) + Bx]/ (x^2 + x)
[A(x + 1) + Bx]/ (x^2 + x) = 1/(x^2+x)
=> Ax + A + Bx = 1
Equate the terms with x and the numeric terms.
A + B = 0
A = 1
B = -1
Therefore we get (1/x) - (1/(x +1))
Given the result of addition 1/(x^2+x) what are the two elementary fractions.
We know the denominator x^2+x = x(x+1).
So the given rational function1/x^2+x = 1/x(x+1).
Let two simple fractions be A/x+B/x+1 whose sum is 1/x(x+1).
So A/x+B/x+1 = 1/x(x+1).
Mutipy both sides by x(x+1).
A(x+1)+Bx = 1.
Ax+A+Bx = 1.
(A+B)x+B = 1.
Equating the contant terms, we get A = 1.
Equating the coefficient of x terms, A+B = 0. 1+B = 0. So B = 1.
Therefore 1/x^2+x = (1/x) + (-1/(x+1)).
So addition of 1/x and -1/(x+1) gives 1/(x^2+x).
First, we'll factorize by x the denominator:
1/(x^2+x) = 1/x(x + 1)
We notice that the denominator of the given fraction is the least common denominator of 2 irreducible ratios.
The final ratio 1/x(x + 1) is the result of addition or subtraction of 2 elementary fractions, as it follows:
1/x(x + 1) = A/x + B/(x+1) (1)
We'll multiply by x(x + 1) both sides:
1 = A(x+1) + Bx
We'll remove the brackets:
1 = Ax + A + Bx
We'll factorize by x to the right side:
1 = x(A+B) + A
We'll compare expressions of both sides:
A+B = 0
A = 1
1 + B = 0
B = -1
We'll substitute A and B into the expression (1) and we'll get the algebraic sum of 2 elementary fractions:
1/x(x + 1) = 1/x - 1/(x + 1)