# Elementary fractions.Knowing the result of addition 1/(x^2+x) find the two elementary fractions.

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### 2 Answers

What we need to determine is the partial fractions of 1/(x^2+x)

1/(x^2+x) = 1/[x*(x + 1]

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

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

=> Ax + A + Bx = 1

equating the numeric coefficients A = 1

equating the coefficients of x

=> A + B = 1 => B = -1

**The final result is 1/[x*(x + 1] = 1/x - 1/(x + 1)**

We'll have 2 elementary fractions because we notice 2 factors at denominator.

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

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)**