Question 1121155
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\frac{dx}{\sqrt{4x^2\,-\,1}}]


Let *[tex \Large u\ =\ 2x\ \ \right\ \ dx\ =\ \frac{du}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int\,\frac{du}{\sqrt{u^2\,-\,1}}]


Let *[tex \Large v\ =\ \sec^{-1}(u)\ \ \right\ \ du\ =\ \sec(v)\tan(v)dv]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int\,\frac{\sec(v)\tan(v)}{\sqrt{\sec^2(v)\,-\,1}}dv]


But *[tex \Large \sec^2(\varphi)\ -\ 1\ =\ \tan^2(\varphi)] so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int\,\sec(v)dv]


Expand:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int\,\frac{\sec(v)\(\tan(v)\,+\,\sec(v)\)}{\tan(v)\,+\,\sec(v)}dv]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int\,\frac{\sec(v)\tan(v)\,+\,\sec^2(v)}{\tan(v)\,+\,\sec(v)}dv]



Let *[tex \Large w\ =\ \tan(v)\ +\ \sec(v)\ \ \right\ dv\ =\ \frac{dw}{\sec(v)\tan(v)\,+\,\sec^2(v)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int\,\frac{dw}{w}\ =\ \ln|w|]


Undo *[tex \Large w] substitution


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ln\|\tan(v)\ +\ \sec(v)\|}{2}]


Undo *[tex \Large v] substitution:


*[tex \Large v\ =\ \sec^{-1}(u)]


*[tex \Large \tan\(\sec^{-1}(u)\)\ =\ \sqrt{u^2\ -\ 1}]


*[tex \Large \sec\(\sec^{-1}(u)\)\ =\ u]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ln\|\tan(v)\ +\ \sec(v)\|}{2}\ =\ \frac{\ln\|\sqrt{u^2\ -\ 1}\ +\ u\|}{2}]


Undo *[tex \Large u] substitution


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ln\|\sqrt{4x^2\ -\ 1}\ +\ 2x\|}{2}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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