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

Nov 30, 2016

$\int \frac{{x}^{2} - 6 x - 20}{x + 5} \mathrm{dx} = {\left(x + 5\right)}^{2} / 2 - 16 \left(x + 5\right) + 35 \ln \left(x + 5\right)$

Explanation:

Substitute $x = t - 5$

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

$= \int \frac{\mathrm{dt}}{t} \left({t}^{2} - 10 t + 25 - 6 t + 30 - 20\right) =$

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

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