How do you integrate #int (x-3x^2)/((x-6)(x-2)(x+4)) # using partial fractions?

Question

Isabella Ackerman · Accepted Answer

The answer is #=-51/20ln(∣x-6∣)+5/12ln(∣x-2∣)-13/15ln(∣x+4∣)+C#

Let's perform the decomposition into partial fractions

#(x-3x^2)/((x-6)(x-2)(x+4))=A/(x-6)+B/(x-2)+C/(x+4)#

#=(A(x-2)(x+4)+B(x-6)(x+4)+C(x-6)(x-2))/((x-6)(x-2)(x+4))#

Therefore,

#x-3x^2=A(x-2)(x+4)+B(x-6)(x+4)+C(x-6)(x-2)#

Let #x=6#, #=>#, #-102=40A#, #=>#, #A=-51/20#

Let #x=2#, #=>#, #-10=-24B#, #=>#, #B=5/12#

Let #x=-4#, #=>#, #-52=60C#, #=>#, #C=-13/15#

So,

#(x-3x^2)/((x-6)(x-2)(x+4))=(-51/20)/(x-6)+(5/12)/(x-2)+(-13/15)/(x+4)#

Therefore,

#int((x-3x^2)dx)/((x-6)(x-2)(x+4))=-51/20intdx/(x-6)+5/12intdx/(x-2)-13/15intdx/(x+4)#

#=-51/20ln(∣x-6∣)+5/12ln(∣x-2∣)-13/15ln(∣x+4∣)+C#

Elijah Abram · Answer

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

1. Perform long division if necessary to ensure the degree of the numerator is less than the degree of the denominator.
2. Factor the denominator into linear and irreducible quadratic factors.
3. Express the given fraction as a sum of simpler fractions using partial fraction decomposition.
4. Find the unknown constants by equating coefficients.
5. Integrate each partial fraction term separately.

Given the integral $ \int \frac{x - 3x^2}{(x - 6)(x - 2)(x + 4)} $, let's start by performing partial fraction decomposition:

$$ \frac{x - 3x^2}{(x - 6)(x - 2)(x + 4)} = \frac{A}{x - 6} + \frac{B}{x - 2} + \frac{C}{x + 4} $$

To find the constants $ A $, $ B $, and $ C $, we multiply both sides by the common denominator $ (x - 6)(x - 2)(x + 4) $ and simplify:

$$ x - 3x^2 = A(x - 2)(x + 4) + B(x - 6)(x + 4) + C(x - 6)(x - 2) $$

Next, we equate coefficients for like terms:

For $ x^2 $:
$$ -3 = A + B + C $$

For $ x $:
$$ 1 = -2A - 6B - 6C $$

For constants:
$$ 0 = 8A + 24B - 24C $$

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

Once we have the values of $ A $, $ B $, and $ C $, we substitute them back into the partial fraction decomposition:

$$ \int \frac{x - 3x^2}{(x - 6)(x - 2)(x + 4)} = \int \frac{A}{x - 6} + \frac{B}{x - 2} + \frac{C}{x + 4} $$

Finally, we integrate each partial fraction term separately.