`int (x^3-4x^2-4x+20)/(x^2-5)dx`

To solve, divide the numerator by the denominator (see attached figure).

`= int (x - 4 + x/(x^2-5))dx`

`= int xdx - int4dx + int x/(x^2-5)dx`

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

For the second integral, apply the formula `int adx = ax + C` .

`= x^2/2 - 4x +C + int x/(x^2-5)dx`

For the third integral, apply u-substitution method.

Let

`u = x^2-5`

Differentiate u.

`du=2x dx`

`(du)/2 =xdx`

Plug-in them to the third integral.

`=x^2/2 - 4x + C + int 1/(x^2-5)*xdx`

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

`= x^2/2 - 4x + C + 1/2int 1/u du`

Then, apply the formula `int 1/xdx = ln|x| + ` C.

`=x^2/2-4x + 1/2ln|u| + C`

And, substitute back `u = x^2-5` .

`=x^2/2 - 4x +1/2ln|x^2-5|+C`

**Therefore, `int (x^3-4x^2-4x + 20)/(x^2-5)dx = x^2/2 - 4x + 1/2ln|x^2-5|+C` .**

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