# How do you find the indefinite integral of int (x^2-6x-20)/(x+5)?

Apr 13, 2017

The answer is $= \frac{\left(x + 5\right) \left(x - 27\right)}{2} + 35 \ln \left(| x + 5 |\right) + C$

#### Explanation:

We perform this integral by substitution.

Let $u = x + 5$, $\implies$, $\mathrm{dx} = \mathrm{du}$

$x = u - 5$

and

${x}^{2} - 6 x - 20 = {\left(u - 5\right)}^{2} - 6 \left(u - 5\right) - 20$

$= {u}^{2} - 10 u + 25 - 6 u + 30 - 20$

$= {u}^{2} - 16 u + 35$

Therefore,

$\int \frac{\left({x}^{2} - 6 x - 20\right) \mathrm{dx}}{x + 5}$

$= \int \frac{\left({u}^{2} - 16 u + 35\right) \mathrm{du}}{u}$

$= \int \left(u - 16 + \frac{35}{u}\right)$

$= {u}^{2} / 2 - 16 u + 35 \ln u$

$= {\left(x + 5\right)}^{2} / 2 - 16 \left(x + 5\right) + 35 \ln \left(| x + 5 |\right) + C$

$= \left(x + 5\right) \frac{x + 5 - 32}{2} + 35 \ln \left(| x + 5 |\right) + C$

$= \frac{\left(x + 5\right) \left(x - 27\right)}{2} + 35 \ln \left(| x + 5 |\right) + C$