How do you integrate #int x/(x-6) dx# using partial fractions?

Question

How do you integrate #int x/(x-6)  dx# using partial fractions?

Cameron Alexander · Accepted Answer

Don't need to use partial fractions.

Step 1: divide #(x-6)# into #x# using long division: #x/(x-6)=1+6/(x-6)# Step 2: Integrate #int[1+6/(x-6)]dx=x+6ln(x-6)+C# hope that helped

Axel Alvarez · Answer

undefined To integrate the function $ \frac{x}{x-6} $ using partial fractions, follow these steps:

1. Write the fraction as a sum of simpler fractions using partial fraction decomposition: 
   $ \frac{x}{x-6} = \frac{A}{x-6} + \frac{B}{x} $

2. Multiply both sides by the common denominator $ (x-6) \times x $ to clear the fractions:
   $ x = A \times x + B \times (x-6) $

3. Substitute values for $ x $ that make the denominators zero to solve for $ A $ and $ B $.

- Substitute $ x = 0 $: $ 0 = 0 \times A + B \times (0 - 6) = -6B $
   - Substitute $ x = 6 $: $ 6 = A \times 6 + 0 = 6A $

4. Solve the system of equations:
   $ A = 1 $ and $ B = -1 $

5. Rewrite the original integral with the partial fractions:
   $ \int \frac{x}{x-6} dx = \int \frac{1}{x-6} dx - \int \frac{1}{x} dx $

6. Integrate each term separately:
   $ \int \frac{1}{x-6} dx = \ln|x-6| $ and $ \int \frac{1}{x} dx = \ln|x| $

7. Combine the results:
   $ \int \frac{x}{x-6} dx = \ln|x-6| - \ln|x| + C $
   where $ C $ is the constant of integration.