dx-x-4-x-2-a-2-

Question Number 138296 by bobhans last updated on 12/Apr/21

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

Answered by bemath last updated on 12/Apr/21

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

I = \int \frac{dx}{x^{4} \sqrt{x^{2} - a^{2}}} we do the changement x = ach (t) \Rightarrow

I = \int \frac{ash (t)}{a^{4} {ch}^{4} t (asht)} dt = \frac{1}{a^{4}} \int \frac{dt}{{(\frac{1 + ch (2 t)}{2})}^{2}}

= \frac{4}{a^{4}} \int \frac{dt}{1 + 2 ch (2 t) + {ch}^{2} (2 t)} = \frac{4}{a^{4}} \int \frac{dt}{1 + 2 ch (2 t) + \frac{1 + ch (4 t)}{2}}

= \frac{8}{a^{3}} \int \frac{dt}{2 + 4 ch (2 t) + 1 + ch (4 t)} =_{2 t = u} \frac{8}{a^{3}} \int \frac{du}{2 (3 + 4 ch (u) + ch (2 u))}

= \frac{4}{a^{3}} \int \frac{du}{ch (2 u) + 4 chu + 3} = \frac{4}{a^{3}} \int \frac{du}{\frac{e^{2 u} + e^{- 2 u}}{2} + 4 \frac{e^{u} + e^{- u}}{2} + 3}

= \frac{8}{a^{3}} \int \frac{du}{e^{2 u} + e^{- 2 u} + {4 e}^{u} + {4 e}^{- u} + 6}

=_{e^{u} = y} \frac{8}{a^{3}} \int \frac{dy}{y (y^{2} + y^{- 2} + 4 y + {4 y}^{- 1} + 6)}

= \frac{8}{a^{3}} \int \frac{dy}{y^{3} + y^{- 1} + {4 y}^{2} + 4 + 6 y} = \frac{8}{a^{3}} \int \frac{ydy}{y^{4} + 1 + {4 y}^{3} + 4 y + {6 y}^{2}}

rest to decompose F (y) = \frac{y}{y^{4} + {4 y}^{3} + {6 y}^{2} + 4 y + 1}

\dots . be continued \dots .

Answered by Ñï= last updated on 12/Apr/21

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

x = a \sec θ

I = \int \frac{a \tan θ \sec θ}{a^{5} \sec^{4} θ \tan θ} d θ = \frac{1}{a^{4}} \int \cos^{3} θ d θ = \frac{1}{a^{4}} \int (1 - \sin^{2} θ) d (\sin θ)

= \frac{1}{a^{4}} (\sin θ - \frac{1}{3} \sin^{3} θ) + C

= \frac{1}{a^{4}} (\frac{{(x^{2} - a^{2})}^{1 / 2}}{x} - \frac{{(x^{2} - a^{2})}^{3 / 2}}{3 x^{3}}) + C

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