You need to test if denominator can be factored, hence, you may factor out ` x` , such that:

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

You need to replace `x^2 + x` with its factored form `x(x + 1)` such that:

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

Decomposing in partial fractions yields:

`1/(x(x + 1)) = a/x + b/(x + 1)`

`1 = ax + a + bx => 1 = x(a + b) + a`

Equating the coefficients of like powers yields:

`{(a + b = 0),(a = 1):} => b = -1`

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

**Hence, decomposing the given fraction `1/(x^2 + x)` in partial fractions yields **`1/(x(x + 1)) = 1/x - 1/(x + 1).`

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)