How do you integrate #int (x-3x^2)/((x-6)(x-2)(x-5)) # using partial fractions?

Question

Scarlett Allison · Accepted Answer

#int(x-3x^2)/((x-6)(x-2)(x-5))dx#

= #-121/8ln(x-6)+5/8ln(x-2)+35/4ln(x-5)+c# Let us first convert #(x-3x^2)/((x-6)(x-2)(x-5))# into partial fractions. #(x-3x^2)/((x-6)(x-2)(x-5))hArrA/(x-6)+B/(x-2)+C/(x-5)# #(x-3x^2)/((x-6)(x-2)(x-5))hArr(A(x-2)(x-5)+B(x-6)(x-5)+C(x-6)(x-2))/((x-6)(x-2)(x-5))# = #(A(x^2-7x+10)+B(x^2-11x+30)+C(x^2-8x+12))/((x-6)(x-2)(x-5))# = #(x^2(A+B+C)-x(7A+11B+8C)+(10A+30B+12C))/((x-6)(x-2)(x-5))# Hence #A+B+C=-3#, #7A+11B+8C=-1# and #10A+30B+12C=0# Subtracting #7# times of first equation from second and #10# times first from third equation, we get #4B+C=20# and #20B+C=30# which gives us #B=5/8# and #C=140/8=35/4# and putting these in #A+B+C=-3#, we get #A=-121/8# Hence, #(x-3x^2)/((x-6)(x-2)(x-5))=-121/(8(x-6))+5/(8(x-2))+35/(4(x-5))# and #int(x-3x^2)/((x-6)(x-2)(x-5))dx=int[-121/(8(x-6))+5/(8(x-2))+35/(4(x-5))]dx# = #-121/8ln(x-6)+5/8ln(x-2)+35/4ln(x-5)+c#

Evelyn Agard · Answer

undefined To integrate the rational function $ \frac{{x - 3x^2}}{{(x - 6)(x - 2)(x - 5)}} $ using partial fractions, follow these steps:

1. Factor the denominator $ (x - 6)(x - 2)(x - 5) $.
2. Write the given rational function as a sum of partial fractions with undetermined constants.
3. Equate the original rational function to the sum of partial fractions and clear the denominators.
4. Solve for the undetermined constants.
5. Integrate each partial fraction individually.
6. Combine the results to obtain the final integrated form.

Let's proceed:

1. The denominator factors into $ (x - 6)(x - 2)(x - 5) $.

2. We write the given rational function as a sum of partial fractions:

$$ \frac{{x - 3x^2}}{{(x - 6)(x - 2)(x - 5)}} = \frac{A}{x - 6} + \frac{B}{x - 2} + \frac{C}{x - 5} $$

3. Equating the original rational function to the sum of partial fractions:

$$ x - 3x^2 = A(x - 2)(x - 5) + B(x - 6)(x - 5) + C(x - 6)(x - 2) $$

4. Clearing the denominators:

$$ x - 3x^2 = A(x^2 - 7x + 10) + B(x^2 - 11x + 30) + C(x^2 - 8x + 12) $$

5. Expanding and collecting like terms:

$$ x - 3x^2 = (A + B + C)x^2 - (7A + 11B + 8C)x + (10A + 30B + 12C) $$

Now, equate coefficients:

For $ x^2 $: $ A + B + C = -3 $  
For $ x $: $ -7A - 11B - 8C = 1 $  
For the constant term: $ 10A + 30B + 12C = 0 $

Solving these equations will give us the values of $ A $, $ B $, and $ C $.

6. After finding the values of $ A $, $ B $, and $ C $, integrate each partial fraction individually, and then combine the results to obtain the final integrated form.