We wish to find $$J=\int \frac{dx}{(x^2+2x+3)^2}$$ Let $$I(a)=\int \frac{dx}{(x+1)^2+a^2}=\frac{1}{a} \tan^{-1} \frac{(x+1)}{a}$$ Differentiate w.r.t. $a$ both sides to get $$\int \frac{-2a dx}{((x+1)^2+a^2)^2}=\frac{-1}{a^2}\tan^{-1} \frac{(x+1)}{a}-\frac{(x+1)/a}{(x+1)^2+a^2}.$$ Hence, putting $a=\sqrt{2}$ and simplifying, we get $$J=\frac{1}{4\sqrt{2}}\tan^{-1}\frac{x+1}{\sqrt{2}}+\frac{x+1}{4(x^2+2x+3)}.$$
What are other methods of getting the $J$ integral?
On
Note that $$\left(\frac{x+1}{(x+1)^2+2}\right)’=-\frac1{(x+1)^2+2}+\frac4{[(x+1)^2+2]^2} $$ Integrate both sides to obtain $$J=\frac{x+1}{4[(x+1)^2+2]}+ \frac{1}{4\sqrt{2}}\tan^{-1}\frac{x+1}{\sqrt{2}} +C$$
On
Decompose into partial fractions:
$$\frac1{\left(x^2+2x+3\right)^2} = -\frac18 \left[\frac1{\left(x+1+\sqrt2\,i\right)^2} + \frac1{\left(x+1-\sqrt2\,i\right)^2}\right] + \frac i{8\sqrt2} \left[\frac1{x+1+\sqrt2\,i} - \frac1{x+1-\sqrt2\,i}\right]$$
Integration and simplification yields
$$\frac18 \left[\frac1{x+1+\sqrt2\,i} + \frac1{x+1-\sqrt2\,i}\right] + \frac i{8\sqrt2} \left[\log\left(x+1+\sqrt2 i\right) - \log\left(x+1-\sqrt2\,i\right)\right] + C$$
$$\frac14 \cdot \frac{x+1}{x^2+2x+3} + \frac1{4\sqrt2} \cdot \frac i2 \left[\log\left(\frac i{-i}\right) + \log\left(\frac{1 - i\frac{x+1}{\sqrt2}}{1 + i\frac{x+1}{\sqrt2}}\right)\right] + C$$
$$\frac14\cdot\frac{x+1}{x^2+2x+3} + \frac1{4\sqrt2} \arctan\left(\frac{x+1}{\sqrt2}\right) + C$$
where we use the identity
$$\arctan(z) = \frac i2 \log\left(\frac{1-iz}{1+iz}\right)$$
On
Substituting $$x = \sqrt{2} \tan \theta - 1, \qquad dx = \sqrt{2} \sec^2 \theta \,d\theta$$ transforms the integral to $$\frac{1}{2\sqrt 2} \int \cos^2 \theta \,d\theta = \frac{1}{4 \sqrt 2} (\sin \theta \cos \theta + \theta) + C = \frac{x + 1}{4(x^2 + 2 x + 3)} + \frac{1}{4 \sqrt{2}} \arctan \frac{x + 1}{\sqrt{2}} + C.$$
An analogous method applies to all integrals $$\int \frac{dx}{(a x^2 + b x + c)^k}, \qquad k \in \{1, 2, 3, \ldots\} , \qquad \textrm{where} \qquad b^2 - 4 a c < 0 .$$
On
Let:$$I=\int\;\frac {dx}{x^2+2x+3}$$ using integral by part: $$I=\int\;\frac {dx}{x^2+2x+3}=\frac {x}{x^2+2x+3}+\int\;\frac {2x^2+2x}{(x^2+2x+3)^2}dx$$ $$I=\frac {x}{x^2+2x+3}+\int\;\frac {2(x^2+2x+3)-(2x+2)-4}{(x^2+2x+3)^2}dx$$ $$I=\frac {x}{x^2+2x+3}+2\int\;\frac {1}{x^2+2x+3}dx-\int\;\frac {2x+2}{(x^2+2x+3)^2}dx-4\int\;\frac {1}{(x^2+2x+3)^2}dx$$ $$I=\frac {x}{x^2+2x+3}+2I+\frac {1}{x^2+2x+3}-4\int\;\frac {1}{(x^2+2x+3)^2}dx\\~\\\implies\;4\int\;\frac {1}{(x^2+2x+3)^2}dx=\frac {x+1}{x^2+2x+3}+I\\~\\\implies\;J=\int\;\frac {1}{(x^2+2x+3)^2}dx=\frac {x+1}{4x^2+8x+12}+\frac {1}{4}I$$ and therfore: $$I=\int\;\frac {dx}{x^2+2x+3}=\int\;\frac {dx}{(x+1)^2+2}=\frac {1}{\sqrt {2}}\int\;\frac {\textstyle\frac {1}{\sqrt {2}}dx}{\left({\frac {x+1}{\sqrt {2}}}\right)^2+1}=\frac {1}{\sqrt {2}}\tan^{-1}\left({\frac {x+1}{\sqrt {2}}}\right)$$ finally $$J=\int\;\frac {1}{(x^2+2x+3)^2}dx=\frac {x+1}{4x^2+8x+12}+\frac {1}{4\sqrt {2}}\tan^{-1}\left({\frac {x+1}{\sqrt {2}}}\right)+C$$
On
Integration by parts $$ \begin{aligned} \int \frac{d x}{\left(x^2+2 x+3\right)^2} = & -\frac{1}{2} \int \frac{1}{x+1} d\left(\frac{1}{x^2+2x+3}\right) \\ = & -\frac{1}{2(x+1)\left(x^2+2 x+3\right)}-\frac{1}{2} \int \frac{1}{(x+1)^2} \cdot \frac{1}{x^2+2 x+3} d x \\ = & -\frac{1}{2(x+1)\left(x^2+2 x+3\right)}-\frac{1}{4} \int\left[\frac{1}{(x+1)^2}-\frac{1}{x^2+2 x+3}\right] d x \\ = & -\frac{1}{2(x+1)\left(x^2+3 x+3\right)}+\frac{1}{4(x+1)} +\frac{1}{4 \sqrt{2}} \tan ^{-1}\left(\frac{x+1}{\sqrt{2}}\right)+C \\ = & \frac{x+1}{4\left(x^2+2 x+3\right)}+\frac{1}{4 \sqrt{2}} \tan ^{-1}\left(\frac{x+1}{\sqrt{2}}\right)+C \end{aligned} $$
Here's a solution that can be done with standard calculus II techniques. First note that by completing the square, $$x^2 +2x + 3 = (x+1)^2 + 2.$$ Now let $x+1 = \sqrt{2} \tan u$ so that $dx = \sqrt{2} \sec^2 u \ du.$ Now it's routine. So
\begin{align*} J &= \int \frac{dx}{(x^2+2x+3)^2} \\ &= \int \frac{\sqrt{2} \sec^2 u \ du}{(2 + 2\tan^2 u)^2} \\ &= \frac{\sqrt{2}}{4} \int \frac{\sec^2 u \ du}{\sec^4 u} \\ &= \frac{\sqrt{2}}{4} \int \cos^2 u \ du \\ &= \frac{\sqrt{2}}{4} \left(\frac{1}{2} u + \frac{\sin 2u}{4} \right) + C \\ &= \boxed{\frac{\sqrt{2}}{8} \arctan \left(\frac{x+1}{2} \right) + \frac{1}{4} \left( \frac{x+1}{(x+1)^2 + 2} \right) + C }\\ \end{align*}
which agrees with the previous results.