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What's the integral of 1/(x^2-4)?

The integral of 1/(x^2-4) is (1/4)ln|(x-2)/(x+2)| + C.

To solve this integral, we can use partial fractions. First, we factor the denominator as (x+2)(x-2). Then, we can write 1/(x^2-4) as A/(x+2) + B/(x-2), where A and B are constants.

Multiplying both sides by (x+2)(x-2), we get 1 = A(x-2) + B(x+2). We can solve for A and B by setting x = 2 and x = -2, respectively. This gives us A = 1/4 and B = -1/4.

Now, we can rewrite the integral as (1/4)∫(1/(x+2) - 1/(x-2))dx. We can integrate each term separately using the natural logarithm function:

(1/4)∫(1/(x+2) - 1/(x-2))dx = (1/4)ln|(x+2)/(x-2)| + C.

Therefore, the integral of 1/(x^2-4) is (1/4)ln|(x-2)/(x+2)| + C.

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