How do you integrate #int(x+1)/((2x-4)(x+5)(x-2))# using partial fractions?

Question

Aurora Alvarado · Accepted Answer

First we observe that #int(x+1)/((2x-4)(x+5)(x-2))dx=1/2*int (x+1)/[(x+5)*(x-2)^2]dx#  Hence we have to find constants  A,B,C such as  #(x+1)/[(x+5)*(x-2)^2]=A/(x+5)+B/(x-2)+C/(x-2)^2# We can calculate these constants by giving x three different values for example #x=0,x=-1,x=3#  hence we get #(x+1)/[(x+5)*(x-2)^2]=-2/[49*(x+5)]+2/[49*(x-2)]+3/[14(x-2)^2]# Hence now we have that #int(x+1)/((2x-4)(x+5)(x-2))dx=1/2*int (x+1)/[(x+5)*(x-2)^2]dx=
1/2*[int (-2)/(49*(x+5))dx +int 2/(49(x-2))dx+int 3/[(14)(x-2)^2]dx]=
1/2*[-2/49*ln(x+5)+2/49*ln(x-2)-3/14*(1/(x-2))]+c#

Scarlett Allison · Answer

undefined To integrate $ \frac{x+1}{(2x-4)(x+5)(x-2)} $ using partial fractions, follow these steps:

1. First, factor the denominator $ (2x-4)(x+5)(x-2) $ completely if it's not already factored.

2. Write the fraction in the form of partial fractions: $ \frac{x+1}{(2x-4)(x+5)(x-2)} = \frac{A}{2x-4} + \frac{B}{x+5} + \frac{C}{x-2} $.

3. Clear the fractions by multiplying both sides of the equation by the denominator $ (2x-4)(x+5)(x-2) $.

4. Expand and equate coefficients to solve for $ A $, $ B $, and $ C $.

5. Once you have the values of $ A $, $ B $, and $ C $, rewrite the original integral using the partial fraction decomposition.

6. Now, integrate each term separately.

7. Finally, simplify the result if necessary.