# `int (x-3/x+2 )dx`

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Sometimes formatting errors in posting the question can result in different questions being posed.

Phrasing the question instead as:

`int{x-3}/{x+2}dx` split apart into two integrals

`=int {(x+2) - 2-3}/{x+2}dx`

`=int {x+2}/{x+2}dx -5int 1/{x+2}dx` simplify

`=int dx - 5 int 1/{x+2}dx`

`=x-5 ln|x+2| + C` where C is a constant of integration

**The integral is `x-5ln|x+2|+C` **

To solve this question, we need to use the power rule for each of the first term and last term, then the logarithm rule for the middle term.

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

`=int xdx-3int 1/x dx+2int dx`

`=1/2x^2-3ln|x|+2x + C` where C is the constant of integration

**The integral is `1/2x^2-3ln|x|+2x+C` .**