Separate the fraction 1/(x^2+x) into partial ratios?
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:
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).
1/(x^2+x) = 1/x - 1/(x+1).