Precalculus, Chapter 7, 7.4, Section 7.4, Problem 54

(3x+1)/(2x^3+3x^2)
To decompose this to partial fractions, factor the denominator.
2x^3 + 3x^2 = x^2(2x + 3)
Then, write a fraction for each factor. For the repeated factor x, form a partial fraction for each exponent x from 1 to 2. And assign a variable for each numerators.
A/x , B/x^2 and C/(2x + 3)
Add these three fractions and set it equal to the given fraction.
(3x+1)/(x^2(2x+3)) = A/x + B/x^2+C/(2x+3)
To solve for the values of A, B and C, eliminate the fractions in the equation. So multiply both sides by the LCD.
x^2(2x+3) * (3x+1)/(x^2(2x+3)) = (A/x+B/x^2+C/(2x+3)) *x^2(2x+3)
3x+1=Ax(2x+3) + B(2x +3) + Cx^2
Then, plug-in the roots of the factors.
For the factor x^2, its root is x=0
3*0+1=A*0(2*0+3) + B(2*0+3)+C*0^2
1=3B
1/3=B
For the factor (2x + 3), its root is x=-3/2.
3(-3/2)+1=A(-3/2)(2(-3/2)+3)+B(2(-3/2)+3)+C(-3/2)^2
-7/2=A(-3/2)(0)+ B(0)+9/4C
-7/2=9/4C
-14/9=C
To get the value of A, assign any value to x, and plug-in the values of B and C to:
3x+1=Ax(2x+3) + B(2x +3) + Cx^2
Let x = 1.
3*1+1=A*1(2*1+3) + 1/3(2*1+3)+(-14/9)*1^2
4=5A + 5/3-14/9
4=5A+1/9
4-1/9=5A
35/9=5A
7/9=A
So the partial fraction decomposition of the given rational expression is:
(7/9)/x + (1/3)/x^2 + (-14/9)/(2x+3)
This simplifies to:
7/(9x) + 1/(3x^2)-14/(9(2x+3))

To check, express them with same denominators.
7/(9x) + 1/(3x^2)-14/(9(2x+3))=7/(9x) * (x(2x+3))/(x(2x+3)) + 1/(3x^2)*(3(2x+3))/(3(2x+3)) -14/(9(2x+3))*x^2/x^2
= (14x^2+21x)/(9x^2(2x+3)) + (6x+9)/(9x^2(2x+3)) - (14x^2)/(9x^2(2x+3))
Now that they have same denominators, let's proceed to add/subtract them.
= (14x^2+21x+6x+9-14x^2)/(9x^2(2x+3)) = (27x+9)/(9x^2(2x+3)) = (9(3x+1))/(9x^2(2x+3))= (3x+1)/(x^2(2x+3))
= (3x+1)/(2x^3+3x^2)

Therefore, (3x+1)/(2x^3+3x^2)=7/(9x) + 1/(3x^2)-14/(9(2x+3)) .