How do you find the integral of (6x+1)/(x^2+2x+3)?

Mar 18, 2018

$I = 3 \ln | {x}^{2} + 2 x + 3 | - \frac{5}{\sqrt{2}} {\tan}^{-} 1 \left(\frac{x + 1}{\sqrt{2}}\right) + c$

Explanation:

$I = \int \frac{6 x + 1}{{x}^{2} + 2 x + 3} \mathrm{dx}$

$= \int \frac{6 x + 6}{{x}^{2} + 2 x + 3} \mathrm{dx} - \int \frac{5}{{x}^{2} + 2 x + 3} \mathrm{dx}$

$I = 3 \int \frac{2 x + 2}{{x}^{2} + 2 x + 3} \mathrm{dx} - \int \frac{5}{{x}^{2} + 2 x + 1 + 2} \mathrm{dx}$

$I = 3 \int \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} + 2 x + 3\right)}{{x}^{2} + 2 x + 3} \mathrm{dx} - \int \frac{5}{{\left(x + 1\right)}^{2} + {\left(\sqrt{2}\right)}^{2}} \mathrm{dx}$

$I = 3 \ln | {x}^{2} + 2 x + 3 | - 5 \cdot \frac{1}{\sqrt{2}} {\tan}^{-} 1 \left(\frac{x + 1}{\sqrt{2}}\right) + c$
$I = 3 \ln | {x}^{2} + 2 x + 3 | - \frac{5}{\sqrt{2}} {\tan}^{-} 1 \left(\frac{x + 1}{\sqrt{2}}\right) + c$