Integral of $$$\frac{2}{x \sqrt{x^{2} - 4}}$$$
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Find $$$\int \frac{2}{x \sqrt{x^{2} - 4}}\, dx$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = \frac{1}{x \sqrt{x^{2} - 4}}$$$:
$${\color{red}{\int{\frac{2}{x \sqrt{x^{2} - 4}} d x}}} = {\color{red}{\left(2 \int{\frac{1}{x \sqrt{x^{2} - 4}} d x}\right)}}$$
Let $$$u=\frac{1}{x}$$$.
Then $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{x^{2}} = - du$$$.
The integral can be rewritten as
$$2 {\color{red}{\int{\frac{1}{x \sqrt{x^{2} - 4}} d x}}} = 2 {\color{red}{\int{\left(- \frac{1}{\sqrt{1 - 4 u^{2}}}\right)d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = \frac{1}{\sqrt{1 - 4 u^{2}}}$$$:
$$2 {\color{red}{\int{\left(- \frac{1}{\sqrt{1 - 4 u^{2}}}\right)d u}}} = 2 {\color{red}{\left(- \int{\frac{1}{\sqrt{1 - 4 u^{2}}} d u}\right)}}$$
Let $$$u=\frac{\sin{\left(v \right)}}{2}$$$.
Then $$$du=\left(\frac{\sin{\left(v \right)}}{2}\right)^{\prime }dv = \frac{\cos{\left(v \right)}}{2} dv$$$ (steps can be seen »).
Also, it follows that $$$v=\operatorname{asin}{\left(2 u \right)}$$$.
Thus,
$$$\frac{1}{\sqrt{1 - 4 u ^{2}}} = \frac{1}{\sqrt{1 - \sin^{2}{\left( v \right)}}}$$$
Use the identity $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:
$$$\frac{1}{\sqrt{1 - \sin^{2}{\left( v \right)}}}=\frac{1}{\sqrt{\cos^{2}{\left( v \right)}}}$$$
Assuming that $$$\cos{\left( v \right)} \ge 0$$$, we obtain the following:
$$$\frac{1}{\sqrt{\cos^{2}{\left( v \right)}}} = \frac{1}{\cos{\left( v \right)}}$$$
Integral can be rewritten as
$$- 2 {\color{red}{\int{\frac{1}{\sqrt{1 - 4 u^{2}}} d u}}} = - 2 {\color{red}{\int{\frac{1}{2} d v}}}$$
Apply the constant rule $$$\int c\, dv = c v$$$ with $$$c=\frac{1}{2}$$$:
$$- 2 {\color{red}{\int{\frac{1}{2} d v}}} = - 2 {\color{red}{\left(\frac{v}{2}\right)}}$$
Recall that $$$v=\operatorname{asin}{\left(2 u \right)}$$$:
$$- {\color{red}{v}} = - {\color{red}{\operatorname{asin}{\left(2 u \right)}}}$$
Recall that $$$u=\frac{1}{x}$$$:
$$- \operatorname{asin}{\left(2 {\color{red}{u}} \right)} = - \operatorname{asin}{\left(2 {\color{red}{\frac{1}{x}}} \right)}$$
Therefore,
$$\int{\frac{2}{x \sqrt{x^{2} - 4}} d x} = - \operatorname{asin}{\left(\frac{2}{x} \right)}$$
Add the constant of integration:
$$\int{\frac{2}{x \sqrt{x^{2} - 4}} d x} = - \operatorname{asin}{\left(\frac{2}{x} \right)}+C$$
Answer
$$$\int \frac{2}{x \sqrt{x^{2} - 4}}\, dx = - \operatorname{asin}{\left(\frac{2}{x} \right)} + C$$$A