Find the indefinite integral using integration by partial fractions: S2x/((x^2)-4 dx

This is what I have figured:

2x/[(x-2)(x+2)] = A/(x-2) + B/(x+2)

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

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

-4=-4B So B=1 then let x=2

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

4=4A So A=1

so S(1/(x-2) +1/(x+2))dx

S(1/x-2)dx + S(1/x+2)dx

ln|x-2|+ln|x+2|+C

please let me know if I am on the right track with this one.

Thanks

### 1 Answer | Add Yours

`(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.

**Sources:**

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