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

To solve, divide the numerator by the denominator.

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

Express it as sum of two integrals.

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

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

`= x^3/3 + C + int 5/(x-3)dx`

For the second integral, apply u-substitution method.

Let

`u = x-3`

Differentiate the u.

`du = dx`

Then, plug-in them to the second integral.

`=x^3/3+C +5 int 1/(x-3)dx`

`=x^3/3+C + 5int1/udu`

Apply the integral formula `int 1/xdx = ln|x| + C` .

`= x^3/3 + 5ln|u| + C`

And, substitute back `u = x - 3` .

`= x^3/3+ 5ln|x-3| + C`

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

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