How do you find the integral of int(2x)/(x^2+6x+13) dx

Feb 16, 2015

The answer is: $\ln \left({x}^{2} + 6 x + 13\right) - 3 \arctan \left(\frac{1}{2} \left(x + 3\right)\right) + c$

$\int \frac{2 x}{{x}^{2} + 6 x + 13} \mathrm{dx} = \int \frac{2 x + 6 - 6}{{x}^{2} + 6 x + 13} \mathrm{dx} =$

$= \int \frac{2 x + 6}{{x}^{2} + 6 x + 13} \mathrm{dx} - 6 \int \frac{\mathrm{dx}}{{x}^{2} + 6 x + 13} = \left(1\right)$

${x}^{2} + 6 x + 13 = {x}^{2} + 6 x + 9 + 4 = {\left(x + 3\right)}^{2} + 4 =$

$= 4 \left[\frac{1}{4} {\left(x + 3\right)}^{2} + 1\right] = 4 \left[{\left(\frac{1}{2} \left(x + 3\right)\right)}^{2} + 1\right]$.

So:

$\left(1\right) = \ln \left({x}^{2} + 6 x + 13\right) - 6 \int \frac{\mathrm{dx}}{4 \left[{\left(\frac{1}{2} \left(x + 3\right)\right)}^{2} + 1\right]} =$

$= \ln \left({x}^{2} + 6 x + 13\right) - \frac{6}{4} \int \frac{\mathrm{dx}}{{\left(\frac{1}{2} \left(x + 3\right)\right)}^{2} + 1} =$

$= \ln \left({x}^{2} + 6 x + 13\right) - \frac{3}{2} \cdot 2 \int \frac{\frac{1}{2}}{{\left(\frac{1}{2} \left(x + 3\right)\right)}^{2} + 1} \mathrm{dx} =$

$= \ln \left({x}^{2} + 6 x + 13\right) - 3 \arctan \left(\frac{1}{2} \left(x + 3\right)\right) + c$