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

Question

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

Christopher Agresta · Accepted Answer

#int x^2/((x-1)(x+2))dx =x -4/3lnabs(x+2) +1/3 ln abs(x-1)+C#

As the numerator has the same degree of the denominator we start by separating the rational function in a way that allows a simplification:

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

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

#x^2/((x-1)(x+2)) = color(blue)((x+1)/(x+2)) + 1/((x-1)(x+2)) #

The first addendum can be separated again to simplify:

#x^2/((x-1)(x+2)) = color(blue)((x+2-1)/(x+2)) + 1/((x-1)(x+2)) #

#x^2/((x-1)(x+2)) = color(blue)(1-1/(x+2)) + color(red)(1/((x-1)(x+2)) )#

Now we write the second term (in red) as a sum of partial fractions with parametric numerators:

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

compute the sum in the second member:

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

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

As the denominators are equal, the equation is satisfied when the numerators are equal:

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

which implies:

#{(A+B=0),(2A-B=1):}#

#{(A=-B),(3A=1):}#

#{(A=1/3),(B=-1/3):}#

Substituting in the expressions above:

#x^2/((x-1)(x+2)) = 1-1/(x+2) + 1/3(1/(x-1))-1/3(1/(x+2))#

#x^2/((x-1)(x+2)) = 1 -4/3 (1/(x+2)) + 1/3 (1/(x-1))#

We are now ready to integrate:

#int x^2/((x-1)(x+2))dx = int dx -4/3 int (dx)/(x+2) + 1/3 int(dx)/(x-1)#

#int x^2/((x-1)(x+2))dx =x -4/3lnabs(x+2) +1/3 ln abs(x-1)+C#

Scarlett Allison · Answer

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

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

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

3. Clear the fractions by multiplying both sides by the common denominator $ (x-1)(x+2) $:
   $ x^2 = A(x+2) + B(x-1) $

4. Expand and collect like terms:
   $ x^2 = Ax + 2A + Bx - B $
   $ x^2 = (A + B)x + (2A - B) $

5. Equate the coefficients of like terms:
   $ A + B = 0 $     (coefficients of x)
   $ 2A - B = 1 $    (constant terms)

6. Solve the system of equations:
   From equation (1), $ B = -A $
   Substitute into equation (2): $ 2A - (-A) = 1 $
   $ 3A = 1 $
   $ A = \frac{1}{3} $
   $ B = -\frac{1}{3} $

7. Rewrite the original fraction in terms of the partial fractions:
   $ \frac{x^2}{(x-1)(x+2)} = \frac{1/3}{x-1} - \frac{1/3}{x+2} $

8. Integrate each term separately:
   $ \int \frac{1/3}{x-1} \, dx = \frac{1}{3} \ln|x-1| + C_1 $
   $ \int -\frac{1/3}{x+2} \, dx = -\frac{1}{3} \ln|x+2| + C_2 $

Combine the results:
$$ \int \frac{x^2}{(x-1)(x+2)} \, dx = \frac{1}{3} \ln|x-1| - \frac{1}{3} \ln|x+2| + C $$
Where $ C = C_1 + C_2 $.