How do you integrate #1 / [ (1-2x)(1-x) ]# using partial fractions?

Question

Aurora Alvarado · Accepted Answer

The answer is #=-ln(|1-2x|)+ln(|1-x|)+C#

Now let's begin breaking down into partial fractions.

#1/((1-2x)(1-x))=A/(1-2x)+B/(1-x)#

#=(A(1-x)+B(1-2x))/((1-2x)(1-x))#

We compare the numerators since the denominators are the same.

#1=A(1-x)+B(1-2x)#

Let #x=1#, #=>#, #1=-B#, #=>#, #B=-1#

Let #x=1/2#, #=>#, #1=1/2A#, #=>#, #A=2#

Consequently,

#1/((1-2x)(1-x))=(2)/(1-2x)-(1)/(1-x)#

so,

#intdx/((1-2x)(1-x))=2intdx/(1-2x)-1intdx/(1-x)#

#=2ln(|1-2x|)/(-2)+ln(|1-x|)+C#

Christopher Agresta · Answer

undefined To integrate $ \frac{1}{(1-2x)(1-x)} $ using partial fractions, follow these steps:

1. Decompose the fraction into partial fractions.
2. Find the constants for each partial fraction.
3. Integrate each partial fraction separately.

The steps:

1. Decompose $ \frac{1}{(1-2x)(1-x)} $ into partial fractions.
$$ \frac{1}{(1-2x)(1-x)} = \frac{A}{1-2x} + \frac{B}{1-x} $$

2. Find the constants $ A $ and $ B $.
$$ 1 = A(1-x) + B(1-2x) $$

3. Solve for $ A $ and $ B $ by equating coefficients of like terms.
$$ 1 = (A + B) - (A + 2B)x $$

4. Equate coefficients:
   - Constant term: $ A + B = 1 $
   - Coefficient of $ x $: $ -A - 2B = 0 $

5. Solve the system of equations for $ A $ and $ B $.
$$ A = \frac{2}{3}, \quad B = \frac{1}{3} $$

6. Rewrite the original integral using the partial fractions:
$$ \int \frac{1}{(1-2x)(1-x)} dx = \int \left( \frac{2}{3(1-2x)} + \frac{1}{3(1-x)} \right) dx $$

7. Integrate each partial fraction separately:
$$ = \frac{2}{3} \int \frac{1}{1-2x} dx + \frac{1}{3} \int \frac{1}{1-x} dx $$

8. Perform the integrals:
$$ = -\frac{1}{3} \ln|1-2x| - \frac{1}{3} \ln|1-x| + C $$
$$ = -\frac{1}{3} \ln|1-2x|(1-x)| + C $$

So, the integral of $ \frac{1}{(1-2x)(1-x)} $ using partial fractions is $ -\frac{1}{3} \ln|1-2x||1-x| + C $, where $ C $ is the constant of integration.