Precalculus, Chapter 7, 7.4, Section 7.4, Problem 18

3/(x^2-3x)
To decompose a fraction, factor the denominator.
3/(x(x-3))
Then, write a fraction for each factor. Since the numerator is still unknown, assign a variable to the numerator of each fraction.
A/x and B/(x-3)
Add these two fractions and set it equal to the given rational expression.
3/(x(x-3)) = A/x + B/(x-3)
To get the values of A and B, eliminate the fractions in the equation. So, multiply both sides by the LCD.
x(x-3)*3/(x(x-3))=(A/x+B/(x-3))*x(x-3)
3=A(x-3) + Bx
Then, plug-in the roots of each factor.
For the factor (x-3), its root is x=3.
3=A(3-3) + B(3)
3=3B
3/3=(3B)/3
1=B
For the factor x, its root is x=0.
3=A(0-3)+B(0)
3=-3A
3/(-3)=(-3A)/(-3)
-1=A
So the given rational expression decomposes to:
-1/x + 1/(x-3)
This can be re-written as:
1/(x-3) - 1/x

To check, express the two fractions with same denominators.
1/(x-3)-1/x=1/(x-3)*x/x - 1/x*(x-3)/(x-3)=x/(x(x-3)) - (x-3)/(x(x-3))
Now that they have same denominators, proceed to subtract them.
=(x-(x-3))/(x(x-3)) = (x - x + 3)/(x(x-3))=3/(x(x-3))=3/(x^2-3x)

Therefore, 3/(x^2-3x) = 1/(x-3)-1/x .

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