How do you integrate #int (x^3 + 2x - 1) / (2x^2 - 3x - 2)# using partial fractions?

Question

Isaiah Allen · Accepted Answer

#x^2/4+(3x)/4+17/40ln |x+1/2|+11/5ln|x-2|+const.#

First, we need to obtain a polynomial in the numerator that is of a lower grade than the one in the denominator.

Through extended division:

#" "x^3+0x^2+2x-1" "#|#" "2x^2-3x-2# #-x^3+3/2x^2+x" "#|____ _____#" "1/2x+3/4# #" "3/2x^2+3x-1# #" "-3/2x^2+9/4x+3/2# #" "# _______ #" "21/4x+1/2=1/4(21x+2)#

So the expression becomes #=int(x/2+3/4)dx+1/4int (21x+2)/(2x^2-3x-2)dx#

Let's deal with the last part Finding the zeros of the denominator #2x^2-3x-2=0#

#Delta=9+16=25# => #sqrt(Delta)=5# #x=(3+-5)/4# => #x_1=-1/2#; #x_2=2#

Then we can break the second integrand in this way: #(21x+2)/(2x^2-3x-2)=A/(x+1/2)+B/(x-2)# To find #A# and #B# let's make #x=0 and -1#

#-1=2A-B/2# #-19/3=-2A-B/3# Summing these 2 expressions we get #-22/3=-5/6B# => #B=44/5# #-> A=(-1+B/2)/2=(-1+22/5)/2# => #A=17/10#

So the main expression becomes #=x^2/4+(3x)/4+(1/4)(17/10int dx/(x+1/2)+44/5int dx/(x-2))# #=x^2/4+(3x)/4+17/40ln |x+1/2|+11/5ln|x-2|+const.#

Samantha Adkison · Answer

undefined To integrate the function $ \frac{{x^3 + 2x - 1}}{{2x^2 - 3x - 2}} $ using partial fractions, follow these steps:

1. Factor the denominator $2x^2 - 3x - 2$ into linear factors.
   $2x^2 - 3x - 2 = (2x + 1)(x - 2)$

2. Write the fraction as a sum of partial fractions with undetermined constants.
   $ \frac{{x^3 + 2x - 1}}{{2x^2 - 3x - 2}} = \frac{A}{2x + 1} + \frac{B}{x - 2}$

3. Multiply both sides by the denominator to clear the fractions.
   $x^3 + 2x - 1 = A(x - 2) + B(2x + 1)$

4. Expand and equate coefficients of like terms.
   $x^3 + 2x - 1 = Ax - 2A + 2Bx + B$

5. Group like terms and equate coefficients.
   $x^3 + 2x - 1 = (A + 2B)x + (-2A + B)$

6. Equate coefficients:
   $A + 2B = 1$ (for the coefficient of $x^3$)
   $-2A + B = 2$ (for the constant term)

7. Solve the system of equations to find the values of $A$ and $B$.

8. Once you find the values of $A$ and $B$, substitute them back into the partial fraction decomposition.

9. Integrate each term separately.

10. Finally, sum up the integrals to get the result.

Isabella Ackerman · Answer

undefined To integrate $ \int \frac{x^3 + 2x - 1}{2x^2 - 3x - 2} $ using partial fractions, first factorize the denominator. It factors as $ (2x + 1)(x - 2) $. Then, you decompose the rational function into partial fractions:

$$ \frac{x^3 + 2x - 1}{(2x + 1)(x - 2)} = \frac{A}{2x + 1} + \frac{B}{x - 2} $$

Next, you find the values of $ A $ and $ B $ by multiplying both sides by the denominator and comparing coefficients. After finding the values of $ A $ and $ B $, you integrate each term separately.