## Calculus (3rd Edition)

$$-\frac{1}{6}\left[\frac{(x+3)}{\left(x^{2}+6 x+6\right)}+\frac{1}{\sqrt{3}} \ln \left|\frac{x+3-\sqrt{3}}{\sqrt{x^{2}+6 x+6}}\right|\right]+C$$
Given $$\int \frac{d x}{\left(x^{2}+6 x+6\right)^{2}}$$ Since \begin{align*} x^{2}+6 x+6&=(x+3)^{2}-9 +6\\ &=(x+3)^{2}-3 \end{align*} Let $$x+3=\sqrt{3} \sec u\ \ \ \ \ \ dx=\sqrt{3}\sec u\tan u du$$ then \begin{align*} \int \frac{d x}{\left(x^{2}+6 x+6\right)^{2}}&=\int \frac{d x}{[(x+3)^{2}-3]^2}\\ &= \int \frac{\sqrt{3}\sec u\tan u du}{[3\sec^2 u-3]^2}\\ &= \frac{1}{9}\int \frac{\sqrt{3}\sec u\tan u du}{ \tan^4u}\\ &=\frac{1}{3\sqrt{3}}\int \cot^2u \csc udu\\ &= \frac{1}{3\sqrt{3}}\int (\csc^2u -1)\csc udu\\ &=\frac{1}{3\sqrt{3}}\int (\csc^3u -\csc u)\csc udu\\ \end{align*} Use $$\int \csc ^{3} u d u=-\frac{1}{2} \csc u \cot u+\frac{1}{2} \ln |\csc u-\cot u|+C$$ Hence \begin{align*} \int \frac{d x}{\left(x^{2}+6 x+6\right)^{2}} &=\frac{1}{3\sqrt{3}}\int (\csc^3u -\csc u)\csc udu\\ &=\frac{1}{3\sqrt{3}}\left( -\frac{1}{2} \csc u \cot u+\frac{1}{2} \ln |\csc u-\cot u|+\frac{1}{2} \ln |\csc u-\cot u|+C\right)\\ &= \frac{1}{3\sqrt{3}}\left( -\frac{1}{2} \csc u \cot u+ \ln |\csc u-\cot u|+C\right)\\ &=-\frac{1}{6 \sqrt{3}}\left[\frac{\sqrt{3}}{\sqrt{u^{2}-3}} \cdot \frac{u}{\sqrt{u^{2}-3}}+\ln \left|\frac{u}{\sqrt{u^{2}-3}}-\frac{\sqrt{3}}{\sqrt{u^{2}-3}}\right|\right]+C\\ &=-\frac{1}{6}\left[\frac{(x+3)}{\left(x^{2}+6 x+6\right)}+\frac{1}{\sqrt{3}} \ln \left|\frac{x+3-\sqrt{3}}{\sqrt{x^{2}+6 x+6}}\right|\right]+C \end{align*}