`(2x)/(x^2-4) = (2x)/((x-2)(x+2))`

Using partial fractions;

`(2x)/((x-2)(x+2)) = A/(x-2)+B(x+2)`

`2x = A(x+2)+B(x-2)`

`2x = x(A+B)+2(A-B)`

When you consider x; in the left side the component of x (the number with x term) is 2 and the right side it is (A+B).

So;

`2 = A+B -----(1)`

When you consider constant term ; in the left side the component of constant is 0 and the right side it is 2(A-B).

`0 = 2(A-B) `

`0 = A-B ----(2)`

Once you solve (1) and (2);

`A = 1`

`B = 1`

So we can write;

`(2x)/((x-2)(x+2)) = 1/(x-2)+1(x+2)`

`int (2x)/((x-2)(x+2))dx`

`= int1/(x-2)+1(x+2) dx`

`= int 1/(x-2)dx+int1/(x+2)dx`

`= ln(x-2)+ln(x+2)+C` C is a constant.

`= ln((x-2)(x+2))+C`

**So**

`int (2x)/((x-2)(x+2))dx = ln((x-2)(x+2))+C`

Note:

Your method is correct. This is another way that may be easy to use with complex situations.

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