How do you integrate #int frac{2x-4}{(x-4)(x+3)(x-6)} dx# using partial fractions?

Question

Samantha Adkison · Accepted Answer

#int(2x-4)/((x-4)(x+3)(x-6))dx#

#= -2/7ln|x-4| - 10/63ln|x+3| + 4/9ln|x-6|+C# Applying partial fraction decomposition: #(2x-4)/((x-4)(x+3)(x-6)) = A/(x-4)+B/(x+3)+C/(x-6)# #=> 2x-4 = A(x+3)(x-6) + B(x-4)(x-6) + C(x-4)(x+3)# #=> 2x-4 = (A+B+C)x^2 + (-3A -10B -C)x +(-18A +24B -12C)# Equating equivalent coefficients gives us the system #{(A+B+C = 0), (-3A -10B - C = 2), (-18A +24B -12C = 4):}# Solving, we get #{(A = -2/7), (B=-10/63), (C = 4/9):}# Thus, substituting back in, #(2x-4)/((x-4)(x+3)(x-6)) = -(2/7)/(x-4)-(10/63)/(x+3)+(4/9)/(x-6)# and so, integrating, #int(2x-4)/((x-4)(x+3)(x-6))dx# # = int(-(2/7)/(x-4)-(10/63)/(x+3)+(4/9)/(x-6))dx# # = -(2/7)int1/(x-4)dx -(10/63)int1/(x+3)dx# #+(4/9)int1/(x-6)dx# #= -2/7ln|x-4| - 10/63ln|x+3| + 4/9ln|x-6|+C#

Scarlett Allison · Answer

#int frac{2x - 4}{(x - 4)(x + 3)(x - 6)} dx #
#= -2/7 ln|x - 4| - 10/63 ln|x + 3| + 4/9 ln|x - 6| + C#

where #C# is the constant of integration

First, you need to write out the partial fractions. The denominator has already been factorized for you. #frac{2x - 4}{(x - 4)(x + 3)(x - 6)} -= frac{A}{x - 4} + frac{B}{x + 3} + frac{C}{x - 6}# Where #A#, #B# and #C# are constants to be determined. Note that the sign #-=# means that the equality holds true for all possible values of #x#. Get rid of the denominators by multiplying both sides with #(x - 4)(x + 3)(x - 6)#. #2x - 4 -= A(x + 3)(x - 6) + B(x - 4)(x - 6) + C(x - 4)(x + 3)# When #x = 4# #2(4) - 4 = A(4 + 3)(4 - 6)# #A = -2/7# When #x = -3# #2(-3) - 4 = B(-3 - 4)(-3 - 6)# #B = -10/63# When #x = 6# #2(6) - 4 = C(6 - 4)(6 + 3)# #C = 4/9# Therefore, #frac{2x - 4}{(x - 4)(x + 3)(x - 6)} -= -frac{2/7}{x - 4} - frac{10/63}{x + 3} + frac{4/9}{x - 6}#. Now, we proceed with the integration. #int frac{2x - 4}{(x - 4)(x + 3)(x - 6)} dx = int (-frac{2/7}{x - 4} - frac{10/63}{x + 3} + frac{4/9}{x - 6}) dx# #= -2/7 int frac{1}{x - 4} dx - 10/63 int frac{1}{x + 3} dx + 4/9 int frac{1}{x - 6} dx# #= -2/7 ln|x - 4| - 10/63 ln|x + 3| + 4/9 ln|x - 6| + C#, where #C# is the constant of integration.

Axel Alvarez · Answer

undefined To integrate $ \int \frac{2x - 4}{(x - 4)(x + 3)(x - 6)} \, dx $ using partial fractions, follow these steps:

1. Perform partial fraction decomposition by expressing the rational function as the sum of simpler fractions.
2. Determine the constants in the decomposition.
3. Integrate each term separately.

Let's proceed:

1. Write the partial fraction decomposition:
$$ \frac{2x - 4}{(x - 4)(x + 3)(x - 6)} = \frac{A}{x - 4} + \frac{B}{x + 3} + \frac{C}{x - 6} $$

2. Clear the denominators:
$$ 2x - 4 = A(x + 3)(x - 6) + B(x - 4)(x - 6) + C(x - 4)(x + 3) $$

3. Solve for $ A $, $ B $, and $ C $ by substituting appropriate values of $ x $:
$$ 2x - 4 = A(x^2 - 3x - 6x + 18) + B(x^2 - 4x - 6x + 24) + C(x^2 + 3x - 4x - 12) $$

4. Expand and collect like terms:
$$ 2x - 4 = A(x^2 - 9x + 18) + B(x^2 - 10x + 24) + C(x^2 - x - 12) $$

5. Equate coefficients of corresponding terms:
$$ 2x - 4 = (A + B + C)x^2 + (-9A - 10B - C)x + (18A + 24B - 12C) $$

6. Equate coefficients of corresponding terms:
$$ 2x - 4 = (A + B + C)x^2 + (-9A - 10B - C)x + (18A + 24B - 12C) $$

$$ \begin{cases} 
A + B + C = 0 \
-9A - 10B - C = 2 \
18A + 24B - 12C = -4
\end{cases} $$

7. Solve the system of equations for $ A $, $ B $, and $ C $.

8. Once you find the values of $ A $, $ B $, and $ C $, rewrite the original integral using the partial fractions.

9. Integrate each term separately.

After integrating, you will obtain the result in terms of $ A $, $ B $, and $ C $ along with $ x $, which represents the integral of the given expression.