# Given the result of addition 1/(x^2+x) what are the two elementary fractions?

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### 3 Answers

We have 1/(x^2+x). Now this has to be expressed as the sum of two fractions

1/(x^2+x)

=> 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)**