How do you integrate #f(x)=x/((x-2)(x+4)(x-7))# using partial fractions?

Question

Charles Anctil · Accepted Answer

#I = 1/13ln|x-2| - 18/143ln|x+4| + 7/143ln|x-7| + c#

Let's say that there are constant #a#, #b# and #d# so

#a/(x-2) + b/(x+4) + d/(x-7) = x/((x-2)(x+4)(x-7))#

Solving it back we have

#(a(x+4)(x-7) + b(x-2)(x-7) + d(x-2)(x+4))/((x-2)(x+4)(x-7)) = x/((x-2)(x+4)(x-7))#

That also means,

#ax^2+4ax-7ax+28a+bx^2-9bx+14b+dx^2+2dx-8d = x#

Or

#x^2(a+b+d) + x(4a-7a-9b+2d) + (28a+14b-8d) = x#

Or, still

#a + b + d = 0# #-3a-9b+2d = 1# #14a+7b-4d = 0#

From there, it's just a question of solving the system. Using Cramer's rule for example; solve whichever you way you think is best. For brevity's sake I won't add more to it here, if you think you could benefit from an explanation on that, just contact me.

Anyhow, solving that we see that #a = 1/13#, #b = -18/143# and #d = 7/143#

From there we can just say

#I = 1/13intdx/(x-2) -18/143intdx/(x+4) + 7/143intdx/(x-7)#

Which are easy integrals

#I = 1/13ln|x-2| - 18/143ln|x+4| + 7/143ln|x-7| + c#

Emilia Aguilar · Answer

undefined To integrate $ f(x) = \frac{x}{(x-2)(x+4)(x-7)} $ using partial fractions, follow these steps:

1. Factor the denominator:
   $ (x-2)(x+4)(x-7) $ factors into $ (x-2)(x+4)(x-7) $.

2. Write $ f(x) $ as a sum of partial fractions:
   $ \frac{x}{(x-2)(x+4)(x-7)} = \frac{A}{x-2} + \frac{B}{x+4} + \frac{C}{x-7} $

3. Clear the denominators by multiplying both sides by $ (x-2)(x+4)(x-7) $:
   $ x = A(x+4)(x-7) + B(x-2)(x-7) + C(x-2)(x+4) $

4. Solve for $ A $, $ B $, and $ C $ by substituting values for $ x $ that eliminate two of the terms:
   - Let $ x = 2 $: $ 2 = 6A $ (solving for $ A $)
   - Let $ x = -4 $: $ -4 = -6B $ (solving for $ B $)
   - Let $ x = 7 $: $ 7 = 9C $ (solving for $ C $)

5. Solve for $ A $, $ B $, and $ C $:
   $ A = \frac{1}{3} $, $ B = \frac{2}{3} $, $ C = \frac{7}{9} $

6. Rewrite the original integral with the partial fraction decomposition:
   $ \int \frac{x}{(x-2)(x+4)(x-7)} \, dx = \int \left(\frac{1}{3(x-2)} + \frac{2}{3(x+4)} + \frac{7}{9(x-7)}\right) \, dx $

7. Integrate each term:
   $ \frac{1}{3} \ln|x-2| + \frac{2}{3} \ln|x+4| - \frac{7}{9} \ln|x-7| + C $

8. Combine the terms and add the constant of integration $ C $:
   $ \frac{1}{3} \ln|x-2| + \frac{2}{3} \ln|x+4| - \frac{7}{9} \ln|x-7| + C $