Question

Evaluate the integral
integral of 1/x(sqrt(x^2-4)) dx

Answer 1

no x outside the parentheses in the denominator then?

Answer 2

\[\int\limits\limits1/(x\sqrt{x^2-4})\]

Answer 3

\[\int\limits \frac{1}{x\sqrt{x^2-4}} \, dx=-\frac{1}{2} \text{ArcTan}\left[\frac{2}{\sqrt{-4+x^2}}\right]+C \]

Answer 4

\[\int\frac{dx}{x\sqrt{x^2-4}}\quad;\quad x=2\cosh{y}\quad,\quad dx=2\sinh{y}\,\,dy\]\[\int\frac{dx}{x\sqrt{x^2-4}}=\int\frac{2\sinh{y}\,\,dy}{2\cosh{y}\sqrt{4\cosh^2{y}-4}}=\]\[=\int\frac{2\sinh{y}\,\,dy}{2\cosh{y}\cdot 2\sinh{y}} =\frac{1}{2}\int\frac{dy}{\cosh{y}}=\int\frac{dy}{e^y+e^{-y}}=\]\[\int\frac{e^y}{(e^y)^2+1}dy=\arctan{e^y}+C=\arctan{(\cosh{y}+\sinh{y})}+C=\]\[=\arctan{\left(\frac{x}{2}+\sqrt{\frac{x^2}{4}-1}\right)}+C=\arctan{\frac{x+\sqrt{x^2-4}}{2}}+C\]

Answer 5

Refer to the pdf attachment.