# How do you find the indefinite integral of int (x^2-3x^2+5)/(x-3)?

##### 1 Answer
Nov 22, 2016

Start by simplifying the numerator.

$\implies \int \frac{5 - 2 {x}^{2}}{x - 3}$

We now divide $5 - 2 {x}^{2}$ by $x - 3$.

Hence, our integral becomes $\int \left(- 2 x - 6 - \frac{13}{x - 3}\right)$, which can be integrated using the rule $\int \left({x}^{n}\right) \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$ and the rule $\int \left(\frac{1}{u}\right) \mathrm{du} = \ln | u | + C$.

$\implies - {x}^{2} - 6 x - 13 \ln | x - 3 | + C$

Hopefully this helps!