# Question #0b19a

$\int \frac{\left(2 x + 13\right) \mathrm{dx}}{{x}^{2} + 6 x + 13} =$
$\ln \left({x}^{2} + 6 x + 13\right) + \frac{7}{2} \cdot \arctan \left(\frac{x + 3}{2}\right) + C$

#### Explanation:

From the given $\int \frac{\left(2 x + 13\right) \mathrm{dx}}{{x}^{2} + 6 x + 13}$

We can expand the integrand this way

$\int \frac{\left(2 x + 13\right) \mathrm{dx}}{{x}^{2} + 6 x + 13} = \int \frac{\left(2 x + 6 + 7\right) \mathrm{dx}}{{x}^{2} + 6 x + 13}$

Then we split the fraction

$\int \frac{\left(2 x + 13\right) \mathrm{dx}}{{x}^{2} + 6 x + 13} = \int \frac{\left(2 x + 6\right) \mathrm{dx}}{{x}^{2} + 6 x + 13} + \int \frac{\left(7\right) \mathrm{dx}}{{x}^{2} + 6 x + 13}$

We also know that ${x}^{2} + 6 x + 13 = {\left(x + 3\right)}^{2} + 4$
So that we can also write the integrals this way

$\int \frac{\left(2 x + 13\right) \mathrm{dx}}{{x}^{2} + 6 x + 13} = \int \frac{\left(2 x + 6\right) \mathrm{dx}}{{x}^{2} + 6 x + 13} + 7 \cdot \int \frac{\mathrm{dx}}{{\left(x + 3\right)}^{2} + 4}$

Use now use the formulas

$\int \frac{\mathrm{du}}{u} = \ln u$ and $\int \frac{\mathrm{du}}{{u}^{2} + {a}^{2}} = \frac{1}{a} \cdot \arctan \left(\frac{u}{a}\right)$

it follows

$\int \frac{\left(2 x + 13\right) \mathrm{dx}}{{x}^{2} + 6 x + 13} =$

$\ln \left({x}^{2} + 6 x + 13\right) + \frac{7}{2} \cdot \arctan \left(\frac{x + 3}{2}\right) + C$

God bless.....I hope the explanation is useful.