How do you use partial fraction decomposition to decompose the fraction to integrate #(3x)/((x + 2)(x - 1))#?

Question

Emilia Aguilar · Accepted Answer

The required format in partial fraction is#2/(x+2) + 1/(x-1)#

Let us consider two constants A and B such that #A/(x+2) + B/(x-1)# Now taking L.C.M we get #(A(x-1)+B(x+2))/((x-1)(x+2)) = 3x/((x+2)(x-1))# Comparing the numerators we get #(A(x-1)+B(x+2))=3x# Now putting x=1 we get B=1 And putting x=-2 we get A=2 So required form is #2/(x+2) + 1/(x-1)#

Hope it helps!!

Scarlett Allison · Answer

undefined To decompose the fraction $ \frac{3x}{(x + 2)(x - 1)} $ using partial fraction decomposition, follow these steps:

1. Write the fraction in the form $ \frac{A}{x + 2} + \frac{B}{x - 1} $.
2. Clear the denominators by multiplying both sides of the equation by $ (x + 2)(x - 1) $.
3. Simplify the equation and solve for the constants $ A $ and $ B $.

Here's how it's done:

1. Write the fraction in the form $ \frac{A}{x + 2} + \frac{B}{x - 1} $:
$$ \frac{3x}{(x + 2)(x - 1)} = \frac{A}{x + 2} + \frac{B}{x - 1} $$

2. Clear the denominators:
$$ 3x = A(x - 1) + B(x + 2) $$

3. Expand and simplify:
$$ 3x = Ax - A + Bx + 2B $$
$$ 3x = (A + B)x + (2B - A) $$

Now, since this equation holds for all values of $ x $, the coefficients of corresponding terms must be equal. Thus:

Coefficient of $ x $ on the left side = Coefficient of $ x $ on the right side
$$ 3 = A + B $$

Constant term on the left side = Constant term on the right side
$$ 0 = 2B - A $$

Solve this system of equations to find the values of $ A $ and $ B $. After finding the values, substitute them back into the original partial fraction decomposition.

Once you have $ A $ and $ B $, the decomposition of $ \frac{3x}{(x + 2)(x - 1)} $ would be $ \frac{A}{x + 2} + \frac{B}{x - 1} $, where $ A $ and $ B $ are the constants you found.