`int (2x^2 + 7x - 3)/(x - 2) dx` Find the indefinite integral.

`int (2x^2+7x-3)/(x-2)dx`

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

`= int (2x + 11 + 19/(x-2)) dx`

Express it as sum of three integrals.

`= int 2xdx + int11dx + int 19/(x-2)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` .

`= (2x^2)/2 + 11x + C + int 19/(x-2)dx`

`=x^2+11x+C + int 19/(x-2)dx`

...

`int (2x^2+7x-3)/(x-2)dx`

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

`= int (2x + 11 + 19/(x-2)) dx`

Express it as sum of three integrals.

`= int 2xdx + int11dx + int 19/(x-2)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` .

`= (2x^2)/2 + 11x + C + int 19/(x-2)dx`

`=x^2+11x+C + int 19/(x-2)dx`

For the third integral, use u-substitution method.

Let,

`u = x - 2`

Differentiate u.

`du = dx`

Then, plug-in them to the third integral.

`=x^2+11x+C+19int 1/(x-2)dx`

`=x^2+11x+C+19int 1/udu`

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

`= x^2+11x + 19ln|u| + C`

And substitute back `u = x-2` .

`=x^2+11x+19ln|x-2|+C`

Therefore, `int (2x^2+7x-3)/(x-2)dx = x^2+11x + 19ln|x-2| + C` .