Question

How do you integrate `6/(x^2-25)^2` using partial fractions?

Answer 1

`3/250[ln|(x+5)/(x-5)|-10/(x^2-25)]+C`

Explanation:

`6/(x^2-25)^2 = 6/((x-5)^2(x+5)^2)`

You can carry out the partial fractions decomposition by following the standard method, i.e., by starting with the form

`6/((x-5)^2(x+5)^2) = A/(x-5)+B/(x+5)`
`qquadqquadqquadqquadqquadqquad qquadqquad+C/(x-5)^2+D/(x+5)^2`

and rewriting this as

`6 = A(x-5)(x+5)^2+B(x-5)^2(x+5)`

`qquadqquad +C(x+5)^2+D(x-5)^2`

• substituting `x=+5` gives `100C=6implies C = 3/50`
• substituting `x=-5` gives `100D=6implies D = 3/50`
• comparing coefficients of `x^3` on both sides yield `A +B = 0`
• comparing coefficients of `x^0` on both sides yield
  `-5^3A +5^3B +5^2C+(-5)^2D= 6 implies`
  `A-B = (C+D)/5-6/5^3 = -3/125implies`
  `A = -B = -3/250`

Thus

`6/(x^2-25)^2 = -3/(250(x-5))+3/(250(x+5))`
`qquadqquadqquadqquadqquadqquad qquadqquad+3/(50(x-5)^2)+3/(50(x+5)^2)`

Thus the required integral is

`int (6dx)/(x^2-25)^2 = -int (3dx)/(250(x-5))+int(3dx)/(250(x+5))`
`qquadqquadqquadqquadqquadqquad qquadqquad+int (3dx)/(50(x-5)^2)+int (3dx)/(50(x+5)^2)`
`qquadqquad = 3/250 ln|(x+5)/(x-5)|-3/50[1/(x-5)+1/(x+5)]+C`
`qquadqquad = 3/250[ln|(x+5)/(x-5)|-10/(x^2-25)]+C`