# How do you find the antiderivative of int 1/(x^2+10x+21) dx?

Oct 12, 2016

$\int \frac{1}{{x}^{2} + 10 x + 21} \mathrm{dx} = \frac{1}{4} \ln \left\mid x + 3 \right\mid + \frac{1}{4} \ln \left\mid x + 7 \right\mid + C$

#### Explanation:

$\frac{1}{{x}^{2} + 10 x + 21} = \frac{1}{\left(x + 3\right) \left(x + 7\right)}$

$\textcolor{w h i t e}{\frac{1}{{x}^{2} + 10 x + 21}} = \frac{1}{4} \left(\frac{\left(x + 7\right) - \left(x + 3\right)}{\left(x + 3\right) \left(x + 7\right)}\right)$

$\textcolor{w h i t e}{\frac{1}{{x}^{2} + 10 x + 21}} = \frac{1}{4} \left(\frac{1}{x + 3} - \frac{1}{x + 7}\right)$

So:

$\int \frac{1}{{x}^{2} + 10 x + 21} \mathrm{dx} = \int \frac{1}{4} \left(\frac{1}{x + 3} - \frac{1}{x + 7}\right) \mathrm{dx}$

$\textcolor{w h i t e}{\int \frac{1}{{x}^{2} + 10 x + 21} \mathrm{dx}} = \frac{1}{4} \ln \left\mid x + 3 \right\mid + \frac{1}{4} \ln \left\mid x + 7 \right\mid + C$