Evaluate the integral.

integral 1/((x-2)(x+2))
Note: Use an upper-case "C" for the constant of integration.

I keep getting 1/4[log(2-x)-log(x+2)]+C and the system is saying its wrong... Please help!

I got

(1/4) (ln(x-2) - ln(x+2) ) + c

how did your x-2 from the original denominator turn into 2-x)
I used partial fractions

Ty

To evaluate the given integral, you need to perform partial fraction decomposition. This will involve expressing the rational function 1/((x-2)(x+2)) as a sum of simpler fractions.

Here's how you can do it step by step:

Step 1: Factor the denominator:
(x - 2)(x + 2)

Step 2: Write the partial fraction decomposition:
1/((x - 2)(x + 2)) = A/(x - 2) + B/(x + 2)

Step 3: Determine the values of A and B. To do this, you can use a common denominator and equate the numerators:
1 = A*(x + 2) + B*(x - 2)

Step 4: Solve for A and B. You can do this by substituting values for x that will eliminate one of the terms:
For x = 2: 1 = A*(2 + 2) + B*(2 - 2)
Simplifying the equation gives you: 1 = 4A

For x = -2: 1 = A*(-2 + 2) + B*(-2 - 2)
Simplifying the equation gives you: 1 = -4B

Solving for A and B individually:
From the equation 1 = 4A, you get A = 1/4
From the equation 1 = -4B, you get B = -1/4

Step 5: Substitute the values of A and B back into the partial fraction decomposition:
1/((x - 2)(x + 2)) = 1/4 * (1 / (x - 2)) - 1/4 * (1 / (x + 2))

Step 6: Evaluate the integrals of the decomposed fractions:
∫(1/((x - 2)(x + 2))) dx = ∫(1/4 * (1 / (x - 2)) - 1/4 * (1 / (x + 2))) dx
= 1/4 * ln|x - 2| - 1/4 * ln|x + 2| + C

Therefore, the correct evaluation of the integral is: 1/4 * ln|x - 2| - 1/4 * ln|x + 2| + C, where C is the constant of integration.