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About LinkBacksHow do I integrate (2x + 3)/(x^2 + 2x + 3)
The denominator has complex roots so I can't break it up to integrate by partial fractions.
Well, you can do exactly that, but this is a better method...Originally Posted by necromanzer52
The first integral can be solved by making the substitutionand the second integral can be solved by making the substitution
, the integral becomes
![\displaystyle \begin{align*} \int{\frac{2x+2}{x^2+2x+3}\,dx} + \int{\frac{1}{\left(x+1\right)^2+2}\,dx} &= \int{\frac{1}{u}\,du} + \int{\frac{1}{\left(\sqrt{2}\tan{\theta}\right)^2 + 2} \, \sqrt{2}\sec^2{\theta}\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}\sec^2{\theta}}{2\tan^2{\theta} + 2}\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}\sec^2{\theta}}{2\left(\tan^2{ \theta} + 1\right)}\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}\sec^2{\theta}}{2\sec^2{\theta} }\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}}{2}\,d\theta} \\ &= \ln{|u|} + \frac{\sqrt{2}}{2}\theta + C \\ &= \ln{\left| x^2 + 2x + 3 \right|} + \frac{\sqrt{2}}{2}\arctan{ \left[ \frac{ \sqrt{2} \left( x+1 \right) }{2} \right] } + C \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} \int{\frac{2x+2}{x^2+2x+3}\,dx} + \int{\frac{1}{\left(x+1\right)^2+2}\,dx} &= \int{\frac{1}{u}\,du} + \int{\frac{1}{\left(\sqrt{2}\tan{\theta}\right)^2 + 2} \, \sqrt{2}\sec^2{\theta}\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}\sec^2{\theta}}{2\tan^2{\theta} + 2}\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}\sec^2{\theta}}{2\left(\tan^2{ \theta} + 1\right)}\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}\sec^2{\theta}}{2\sec^2{\theta} }\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}}{2}\,d\theta} \\ &= \ln{|u|} + \frac{\sqrt{2}}{2}\theta + C \\ &= \ln{\left| x^2 + 2x + 3 \right|} + \frac{\sqrt{2}}{2}\arctan{ \left[ \frac{ \sqrt{2} \left( x+1 \right) }{2} \right] } + C \end{align*})
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