`int (4x - 2/(2x+3)^2)dx`

To solve, express it as difference of two integrals.

`= int 4x dx - int 2/(2x+3)^2dx`

Then, apply negative exponent rule `a^(-m)=1/a^m` .

`= int 4xdx - int 2(2x+3)^(-2)dx`

For the second integral, apply the u-substitution method.

`u = 2x + 3`

`du = 2dx`

Expressing...

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`int (4x - 2/(2x+3)^2)dx`

To solve, express it as difference of two integrals.

`= int 4x dx - int 2/(2x+3)^2dx`

Then, apply negative exponent rule `a^(-m)=1/a^m` .

`= int 4xdx - int 2(2x+3)^(-2)dx`

For the second integral, apply the u-substitution method.

`u = 2x + 3`

`du = 2dx`

Expressing the second integral in terms of u variable, it becomes:

`=int 4xdx - int (2x+3)^(-2) * 2dx`

`=int 4xdx - int u^(-2) du`

For both integrals, apply the formula `int x^ndx= x^(n+1)/(n+1)+C` .

`= (4x^2)/2 - u^(-1)/(-1) + C`

`=2x^2 + u^(-1) + C`

`= 2x^2 + 1/u + C`

And, substitute back `u = 2x + 3`

`=2x^2+1/(2x+3)+C`

**Therefore, `int (4x - 2/(2x+3)^2)dx=2x^2+1/(2x+3)+C` .**