dx-x-x-2-4-

Question Number 146215 by mathdanisur last updated on 12/Jul/21

\int \frac{d x}{x \sqrt{x^{2} - 4}} = ?

Answered by Olaf_Thorendsen last updated on 12/Jul/21

F (x) = \int \frac{d x}{x \sqrt{x^{2} - 4}}

F (2 ch u) = \int \frac{2 sh u d u}{2 ch u \sqrt{{4 ch}^{2} u - 4}}

F (2 ch u) = \int \frac{sh u d u}{ch u \times 2 sh u}

F (2 ch u) = \frac{1}{2} \int \frac{d u}{ch u}

F (2 ch u) = \int \frac{e^{u} d u}{e^{2 u} + 1}

F (2 ch u) = \arctan (e^{u}) + C

F (x) = \arctan (e^{argch \frac{x}{2}}) + C

F (x) = \arctan (e^{\ln (\frac{x}{2} + \sqrt{\frac{x^{2}}{4} - 1})}) + C

F (x) = \arctan (\frac{x}{2} + \sqrt{\frac{x^{2}}{4} - 1}) + C

Answered by qaz last updated on 12/Jul/21

\int \frac{dx}{x \sqrt{x^{2} - 4}} = \int \frac{dx}{x^{2} \sqrt{1 - \frac{4}{x^{2}}}} = - \frac{1}{2} \int \frac{d (\frac{2}{x})}{\sqrt{1 - \frac{4}{x^{2}}}} = - \frac{1}{2} \sin^{- 1} \frac{2}{x} + C

Answered by mathmax by abdo last updated on 12/Jul/21

Φ = \int \frac{dx}{x \sqrt{x^{2} - 4}} we do the changement x = \frac{2}{sint} \Rightarrow \frac{dx}{dt} = - \frac{2 cost}{\sin^{2} t}

\Rightarrow Φ = - 2 \int \frac{cost}{\sin^{2} t \times \frac{2}{sint} \sqrt{\frac{4}{\sin^{2} t} - 4}} dt

= - \int \frac{cost}{sint \times 2 \times \frac{cost}{sint}} dt = - \frac{1}{2} \int dt = - \frac{t}{2} C

sint = \frac{2}{x} \Rightarrow t = \arcsin (\frac{2}{x}) \Rightarrow Φ = - \frac{1}{2} \arcsin (\frac{2}{x}) + C

Commented by mathdanisur last updated on 12/Jul/21

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