Calculus (3rd Edition)

$$-\frac{1}{16}\left(\ln \left|\frac{x}{\sqrt{x^{2}-4}}-\frac{2}{\sqrt{x^{2}-4}}\right|+\frac{2}{\sqrt{x^{2}-4}} \cdot \frac{x}{\sqrt{x^{2}-4}}\right)+C$$
Given $$\int \frac{d x}{\left(x^{2}-4\right)^{2}}$$ Let $$x=2\sec u\ \ \ \ \to\ \ \ \ dx=2\sec u\tan udu$$ Then \begin{align*} \int \frac{d x}{\left(x^{2}-4\right)^{2}}&=\int \frac{2\sec u\tan udu}{\left(4\sec^{2}u-4\right)^{2}}\\ &=\int \frac{2\sec u\tan udu}{\left(4\tan^{2}u\right)^{2}}\\ &=\frac{1}{8}\int \frac{ \sec u du}{ \tan^{3}u}\\ &=\frac{1}{8}\int \cot^2 u\csc u du \\ &=\frac{1}{8}\int( \csc^2 u-1)\csc u du \\ &=\frac{1}{8}\int( \csc^3 u-\csc u) du \\ \end{align*} Use $$\int \csc ^{n} u d u=\frac{-1}{n-1} \cot u \csc ^{n-2} u+\frac{n-2}{n-1} \int \csc ^{n-2} u d u$$ Then $$\int \csc ^{3} u d u=-\frac{1}{2} \csc u \cot u+\frac{1}{2} \ln |\csc u-\cot u|+C$$ Hence \begin{align*} \int \frac{d x}{\left(x^{2}-4\right)^{2}} &=\frac{1}{8}\int( \csc^3 u-\csc u) du \\ &=\frac{1}{8}\left(-\frac{1}{2} \csc u \cot u+\frac{1}{2} \ln |\csc u-\cot u|-\ln |\csc u-\cot u|\right)+C\\ &=\frac{-1}{16}\left( \csc u \cot u+ \ln |\csc u-\cot u|\right)+C\\ &=-\frac{1}{16}\left(\ln \left|\frac{x}{\sqrt{x^{2}-4}}-\frac{2}{\sqrt{x^{2}-4}}\right|+\frac{2}{\sqrt{x^{2}-4}} \cdot \frac{x}{\sqrt{x^{2}-4}}\right)+C \end{align*}