Resolve x+62/(3x-1)^2(2x+3) into partial fractions

Answer 1

To resolve x+62/(3x-1)^2(2x+3) into partial fractions, we express it in the form:

x+62/(3x-1)^2(2x+3) = A/(3x-1) + B/(3x-1)^2 + C/(2x+3)

First, we multiply both sides of the equation by the common denominator, (3x-1)^2(2x+3):

x + 62 = A(3x-1)(2x+3) + B(2x+3) + C(3x-1)^2

Now we can solve for the constants A, B, and C by picking values for x and comparing the coefficients of the resulting polynomial equation.

1. Let x = 1/3:

1/3 + 62 = A(0)(5/3) + B(5/3)
63/3 = B * (5/3)
B = 2

2. Let x = -3/2:

-3/2 + 62 = A(3(-3/2)-1)((-3)-3) + 2((-3-3)) - C(3(-3/2)-1)^2
-1/2 + 124/2 = -20A + (-10)2 - 4C
123/2 = -20A -20 -4C

Now we have two equations involving A and C:

3. -20A - 4C = 123/2
4. 3A + 2= C

Solving equation (4) for C: C = 2 - 3A
Substituting C in equation (3):
-20A - 4(2 - 3A) = 123/2
-20A + 12A = 121/2
-8A = 121/2
A = -(121/16)

Substitute A back into equation (4):
C = (3)(-121/16) + 2 = 2 - 363/16 = (32-363)/16 = -331/16

Now we have the constants A, B, and C:

A = -121/16
B = 2
C = -331/16

So the decomposition of x+62/(3x-1)^2(2x+3) into partial fractions is:

(-121/16)/(3x-1) + 2/(3x-1)^2 + (-331/16)/(2x+3)