Calc1.71

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\int {\frac  {{\sqrt  {x^{2}-4}}}{x}}\,dx\,

x=2\sec \theta \,

dx=2\sec \theta \tan \theta \,d\theta \,

\int {\frac  {{\sqrt  {x^{2}-1}}}{x}}\,dx=\int {\frac  {{\sqrt  {4\sec ^{2}\theta -4}}}{2\sec \theta }}(2)\sec \theta \tan \theta \,d\theta =2\int {\sqrt  {\tan ^{2}\theta }}\tan \theta \,d\theta =2\int \tan ^{2}\theta \,d\theta \,

We must use a trig substitution to get this in a form that is more easily integrable.

2\int \tan ^{2}\theta \,d\theta =2\int (\sec ^{2}\theta -1)d\theta =2[\tan \theta -\theta ]+C\,

Now, we know \sec \theta ={\frac  {x}{2}} so \tan \theta ={\frac  {{\sqrt  {x^{2}-4}}}{2}} and \theta =\operatorname{arcsec} {\frac  {x}{2}}. So

\int {\frac  {{\sqrt  {x^{2}-4}}}{x}}\,dx=2\left[{\frac  {{\sqrt  {x^{2}-4}}}{2}}-\operatorname{arcsec} {\frac  {x}{2}}\right]+C={\sqrt  {x^{2}-4}}-2\operatorname{arcsec} {\frac  {x}{2}}+C\,


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