Solve the integral: 5/x(x^2-4) dx
I've done the partial fractions and found B and C, both equaling 5/8; and that's gonna give me 5/8 ln|x^2-4|. I just need help finding A
A = -5/4
where did you get stuck?
I think I got it. You just make x=0 to find A. Thanks.
let 5/x(x^2-4) = A/x + B/(x+2) + C/(x-2)
= [ A(x^2 - 4) + Bx(x-2) + Cx(x+2)]/(x(x^2 - 4))
then A(x^2 - 4) + Bx(x-2) + Cx(x+2) = 5 for all values of x
let x = 0 ----> -4A + 0 + 0 = 5 , so A = -5/4
let x = 2 ----> 0 + 0 + 8C = 5 , so C = 5/8
let x = -2 ---> 0 -8B + 0 = 5, so B = -5/8
∫ 5/x(x^2-4) dx = ∫ ( (-5/4)/x - (5/8)/(x+2) + (5/8) /(x-2) )dx
= (-5/4) lnx - (5/8) ln(x+2) + (5/8) ln (x-2) + c
= (-5/8)(2 lnx - ln(x+2) - ln(x-2) ) + c
To solve for A in the partial fraction decomposition, you can start by expressing the given expression 5/x(x^2-4) as a sum of simpler fractions.
First, factor the denominator x(x^2-4) as x(x+2)(x-2). Then, you can write the integral as:
∫ (5/x(x+2)(x-2)) dx
Next, write the given fraction in partial fraction form. We already know that B and C both equal 5/8. To find A, we can set up the following equation:
5/x(x+2)(x-2) = A/x + (5/8) / (x+2) + (5/8) / (x-2)
To find A, we can multiply both sides of the equation by x(x+2)(x-2) to eliminate the denominators:
5 = A(x+2)(x-2) + 5/8(x)(x-2) + 5/8(x)(x+2)
Simplify the equation:
5 = A(x^2-4) + (5/8)(x^2-2x) + (5/8)(x^2+2x)
Combine like terms:
5 = (A + 5/8 + 5/8)x^2 + (-4A -5/4)x + (6/8) - (6/8)
Comparing the coefficients on both sides, you can set up the following equations:
1) A + 5/8 + 5/8 = 0 (coefficient of x^2 terms)
2) -4A - 5/4 = 0 (coefficient of x terms)
Solving equation 1 for A, you get:
A = -5/8 - 5/8
A = -5/4
Therefore, the value of A is -5/4.