Precalculus, Chapter 7, 7.4, Section 7.4, Problem 19

1/(2x^2+x)
To decompose this into partial fractions, factor the denominator.
1/(x(2x+1))
Write a fraction for each factor. Since the numerators are still unknown, assign a variable to each numerator.
A/x and B/(2x+1)
Add these two fractions and set it equal to the given fraction.
1/(x(2x+1)) = A/x + B/(2x+1)
To solve for the values of A and B, eliminate the fractions in the equation. So, multiply both sides by the LCD.
x(2x+1)*1/(x(2x+1)) = (A/x + B/(2x+1))*x(2x+1)
1=A(2x+1) + Bx
Then, plug-in the roots of the factors.
For the factor 2x + 1, its root is x=-1/2.
1=A(2*(-1/2)+1) + B(-1/2)
1=A(-1+1)+B(-1/2)
1=-1/2B
-2=B
For the factor x, its root is x=0.
1=A(2x+1)+Bx
1=A(2*0+1)+B*0
1=A
So the partial fraction decomposition of the rational expression is:
1/x + (-2)/(2x+1)
And the sign before the second fraction simplifies to:
1/x - 2/(2x+1)

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

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