# Separate the fraction 1/(x^2+x) into partial ratios?

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### 2 Answers

First, we'll factorize by x the denominator:

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

We notice that the denominator of the right side ratio is the least common denominator of 2 irreducible ratios.

We'll suppose that the ratio 1/x(x+1) is the result of addition or subtraction of 2 elementary fractions:

1/x(x+1) = A/x + B/(x+1) (1)

We'll multiply the ratio A/x by (x+1) and we'll multiply the ratio B/(x+1) by x.

1/x(x+1)= [A(x+1) + Bx]/x(x+1)

Since the denominators of both sides are matching, we'll write the numerators, only.

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

If the expressions from both sides are equivalent, the correspondent coefficients are equal.

A+B = 0

A = 1

1 + B = 0

B = -1

We'll substitute A and B into the expression (1):

**1/x(x+1) = 1/x - 1/(x+1)**

To write 1/(x^2+x) in partial fractions.

We know that the denominator x^2+x = x(x+1).

Therefore the partial fractions of 1/x^2+x could be written as: A/x +B/(1+x)

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

1/x(x+1) = A/x+B/(x+1)...(1)

We multiply both side of (1) by x(x+1):

1 = A(x+1)+Bx.

1 = (A+B)x +A...(2)

Equating th like terms on both sides of (2), we get:

A= 1.

So (A+B)x = 0.

A+B = 0.

B = -A = -1.

Therefore the required partial fractions o 1/(x^2+x) are :

1/x - 1/(x+1).

Or

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