SOLUTION: Find integral (dx/sqr(4x^2-1))
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Question 1121155: Find integral (dx/sqr(4x^2-1))
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Let 
Let \ \ \right\ \ du\ =\ \sec(v)\tan(v)dv)
\tan(v)}{\sqrt{\sec^2(v)\,-\,1}}dv)
But \ -\ 1\ =\ \tan^2(\varphi))
so:
dv)
Expand:
\(\tan(v)\,+\,\sec(v)\)}{\tan(v)\,+\,\sec(v)}dv)
\tan(v)\,+\,\sec^2(v)}{\tan(v)\,+\,\sec(v)}dv)
Let \ +\ \sec(v)\ \ \right\ dv\ =\ \frac{dw}{\sec(v)\tan(v)\,+\,\sec^2(v)})
Undo 
substitution
\ +\ \sec(v)\|}{2})
Undo 
substitution:
)
\)\ =\ \sqrt{u^2\ -\ 1})
\)\ =\ u)
So:
\ +\ \sec(v)\|}{2}\ =\ \frac{\ln\|\sqrt{u^2\ -\ 1}\ +\ u\|}{2})
Undo 
substitution
John
My calculator said it, I believe it, that settles it
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