How to decompose in partial fractions 1/(2x^2+4x)?
The fraction `1/(2x^2+4x)` has to be decomposed into partial fractions. First factorize the denominator.
= `2x(x +2)`
Now take two constants a and b
`1/(2x^2+4x) = a/(2x) + b/(x + 2)`
This has to be solved for a and b.
`1/(2x^2+4x) = (a*(x+2) + b*2x)/((2x)(x + 2))`
`1/(2x^2+4x) = (ax + 2a + 2bx)/((2x)(x + 2))`
Equate the terms containing x.
a + 2b = 0
Another equation that can be derived is 2a = 1
a = 1/2
Substituting in a + 2b = 0
b = -1/4
The partial fraction decomposition is `1/(4x) - 1/(4*(x+2))`
First, we'll factorize by 2x the denominator:
1/(2x^2+4x) = 1/2x(x+2)
We'll suppose that the fraction 1/2x(x+2) is the result of addition or subtraction of 2 irreducible partial fractions:
1/2x(x+2) = A/2x + B/(x+2) (1)
We'll multiply the ratio A/2x by (x+2) and we'll multiply the ratio B/(x+2) by 2x.
1/2x(x+2)= [A(x+2) + 2Bx]/2x(x+2)
Since the denominators of both sides are matching, we'll write the numerators, only.
1 = A(x+2) + 2Bx
We'll remove the brackets:
1 = Ax + 2A + 2Bx
We'll factorize by x to the right side:
1 = x(A+2B) + 2A
Comparing, we'll get:
A+2B = 0
2A = 1 => A = 1/2
1/2 + 2B = 0
B = -1/4
We'll substitute A and B into the expression (1):
The result of decomposition into partial fractions is: 1/2x(x+2) = 1/4x - 1/4(x+2)