`int (1/(2x+5)-1/(2x-5))dx`

To solve, express it as difference of two integrals.

`= int 1/(2x+5)dx - int 1/(2x-5)dx`

Then, apply substitution method.

- `u=2x+5`

`du=2dx`

`1/2du=dx`

- `w=2x-5`

`dw=2dx`

`1/2dw=dx`

Expressing the two integrals in terms of u and w, it becomes

`= int 1/u*1/2 du - int 1/w*1/2dw`

`=1/2int1/u du- 1/2 int1/w dw`

To take the integral of these, apply the formula `int 1/x dx = ln|x|+C` .

`= 1/2 ln|u| - 1/2 ln|w|+C`

And, substitute back `u= 2x+5` and `w=2x-5`.

`=1/2ln|2x+5|-1/2ln|2x-5|+C`

**Therefore, `int
(1/(2x+5)-1/(2x-5))dx=1/2ln|2x+5|-1/2ln|2x-5|+C` .**

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