dx-x-2-x-2-1-

Question Number 124532 by liberty last updated on 04/Dec/20

\int \frac{d x}{x^{2} \sqrt{x^{2} - 1}} ?

Answered by bemath last updated on 04/Dec/20

ϑ (x) = \int \frac{d x}{x^{3} \sqrt{1 - x^{- 2}}} = \int \frac{x^{- 3}}{\sqrt{1 - x^{- 2}}} d x

l e t t i n g z = 1 - x^{- 2} \to d z = 2 x^{- 3} d x

ϑ (x) = \frac{1}{2} \int \frac{d z}{\sqrt{z}} = \frac{1}{2} \int z^{- \frac{1}{2}} d z

ϑ (x) = \sqrt{z} + c = \sqrt{1 - \frac{1}{x^{2}}} + c

ϑ (x) = \frac{\sqrt{x^{2} - 1}}{x} + c

Answered by MJS_new last updated on 04/Dec/20

I^{'} m absolutely crazy and I should be kept

away from students \dots

this is no fun to transform but I had lots of

fun transforming it and the integral \dots look

at the integral \dots {isn}^{'} t it s u p e r e a s y ?!

t = \sqrt{2 x (x + \sqrt{x^{2} - 1})} \Leftrightarrow x = \frac{t^{2}}{2 \sqrt{t^{2} - 1}}

\to d x = \frac{t (t^{2} - 2)}{2 {(t^{2} - 1)}^{3 / 2}} d t

\int \frac{d x}{x^{2} \sqrt{x^{2} - 1}} = 8 \int \frac{{(t^{2} - 1)}^{3 / 2}}{t^{4} (t^{2} - 2)} d x =

= ♡ 4 \int \frac{d t}{t^{3}} ♡ = - \frac{2}{t^{2}} = - \frac{1}{x (x + \sqrt{x^{2} - 1})} =

= \frac{\sqrt{x^{2} - 1} - x}{x} = \frac{\sqrt{x^{2} - 1}}{x} - 1 = \frac{\sqrt{x^{2} - 1}}{x} + C

Commented by bemath last updated on 04/Dec/20

h a h a h a . . y o u a r e n o t c r a z y p r o f . b u t

y o u a r e a m a s t e r o f m a t h a n d

s a n t u y . w k w k w k \dots

Answered by Ar Brandon last updated on 04/Dec/20

I = \int \frac{dx}{x^{2} \sqrt{x^{2} - 1}} = \int \frac{dx}{x^{3} \sqrt{1 - \frac{1}{x^{2}}}}

1 - \frac{1}{x^{2}} = u \Rightarrow \frac{2}{x^{3}} dx = du

I = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \sqrt{u} + C = \sqrt{1 - \frac{1}{x^{2}}} + C

Answered by malwan last updated on 04/Dec/20

x = s e c y \Rightarrow d x = s e c y t a n y d y

∴ \int \frac{s e c y t a n y}{{s e c}^{2} y \sqrt{{s e c}^{2} y - 1}} d y = \int \frac{t a n y}{s e c y \sqrt{{t a n}^{2} y}} d y

= \int \frac{d y}{s e c y} = \int c o s y d y = \frac{\sqrt{x^{2} - 1}}{x} + C

Answered by mathmax by abdo last updated on 04/Dec/20

I = \int \frac{dx}{x^{2} \sqrt{x^{2} - 1}} we do the changement x = cht \Rightarrow

I = \int \frac{sht}{{ch}^{2} t sh (t)} dt = 2 \int \frac{dt}{1 + ch (2 t)} =_{2 t = u} 2 \int \frac{du}{2 (1 + chu)}

= \int \frac{du}{1 + \frac{e^{u} + e^{- u}}{2}} = 2 \int \frac{du}{2 + e^{u} + e^{- u}} =_{e^{u} = y} 2 \int \frac{dy}{y (2 + y + y^{- 1})}

= 2 \int \frac{dy}{2 y + y^{2} + 1} = 2 \int \frac{dy}{{(y + 1)}^{2}} = - \frac{2}{y + 1} + C we have

y = e^{u} = e^{2 t} and t = argch (x) = \ln (x + \sqrt{x^{2} - 1}) \Rightarrow

e^{2 t} = {(x + \sqrt{x^{2} - 1})}^{2} \Rightarrow I = \frac{- 2}{{(x + \sqrt{x^{2} - 1})}^{2} + 1} + C