Question

Question #a0e4b

Answer 1

`(1/4)ln(abs(x-2))-(1/4)(ln(abs(x+2)))+C`

Explanation:

I will assume you meant `int (dx/(x^2-4))`:

Rewrite `x^2-4` as `(x-2)(x+2)`:

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

Rewrite the fraction as `A/(x-2) + B/(x+2)`. Solve for `A` and `B` by multiplying the three fractions by `(x-2)(x+2)`:

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

Plug in `2` for `x` to solve for `A`:

`1=A(4)`

`A=1/4`

Plug in `-2` for `x` to solve for `B`:

`1=-4B`

`B=-1/4`

Now we have:

`int (1/4)(1/(x-2)) + (-1/4)((1/(x+2)))`

This, we can solve by integrating:

`(1/4)ln(abs(x-2))-(1/4)(ln(abs(x+2)))+C`

If you want, you can write this as one log with log rules:

`(1/4)ln(abs((x-2)/(x+2)))+C`