∫ ((ln(x+(√(x^2 −1))))/(√((x^2 −1)^3 ))) dx ?


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Question Number 93193 by john santu last updated on 11/May/20

$\int \frac{\ln (x + \sqrt{x^{2} - 1})}{\sqrt{{(x^{2} - 1)}^{3}}} dx ?$
Commented by abdomathmax last updated on 14/May/20

$I = \int \frac{l n (x + \sqrt{x^{2} - 1})}{\sqrt{{(x^{2} - 1)}^{3}}} d x v h a n g e m e n t x = c h t g i v e$ $I = \int \frac{l n (c h t + s h t)}{\sqrt{{s h}^{6} t}} s h (t) d t$ $= \int \frac{t}{{s h}^{2} t} d t = \int \frac{2 t}{c h (2 t) - 1} d t$ $=_{2 t = u} \int \frac{u}{c h (u) - 1} \frac{d u}{2} = \frac{1}{2} \int \frac{u}{\frac{e^{u} + e^{- u}}{2} - 1} d u$ $= \int \frac{u}{e^{u} + e^{- u} - 2} d u =_{e^{u} = z} \int \frac{l n (z)}{z + z^{- 1} - 2} \times \frac{d z}{z}$ $= \int \frac{l n z}{z^{2} - 2 z + 1} d z = \int \frac{l n z}{{(z - 1)}^{2}} d x$ $= - \frac{1}{z - 1} l n (z) + \int \frac{1}{z - 1} \times \frac{d z}{z}$ $= - \frac{1}{z - 1} l n z + \int (\frac{1}{z - 1} - \frac{1}{z}) d z$ $= - \frac{1}{z - 1} l n z + l n ∣ \frac{z - 1}{z} ∣ + C$ $= - \frac{u}{e^{u} - 1} + l n ∣ \frac{e^{u} - 1}{e^{u}} ∣ + C$ $u = \frac{t}{2} = \frac{1}{2} a r g c h (x) = \frac{1}{2} l n (x + \sqrt{x^{2} - 1}) \Rightarrow$ $e^{u} = \sqrt{x + \sqrt{x^{2} - 1}} \Rightarrow$ $I = - \frac{l n (x + \sqrt{x^{2} - 1})}{2 (\sqrt{x + \sqrt{x^{2} - 1})}} + l n ∣ \frac{\sqrt{x + \sqrt{x^{2} - 1}} - 1}{\sqrt{x + \sqrt{x^{2} - 1}}} ∣ + C$
	Answered by MJS last updated on 11/May/20

	$\int \frac{\ln (x + \sqrt{x^{2} - 1})}{{(x^{2} - 1)}^{3 / 2}} d x =$ $[t = x + \sqrt{x^{2} - 1} \to d x = \frac{\sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}} d t]$ $= 4 \int \frac{t \ln t}{{(t^{2} - 1)}^{2}} d t =$ $by parts$ $u = \ln t \to u^{'} = \frac{1}{t}$ $v^{'} = \frac{4 t}{{(t^{2} - 1)}^{2}} \to v = - \frac{2}{t^{2} - 1}$ $= - \frac{2 \ln t}{t^{2} - 1} + 2 \int \frac{d t}{t (t^{2} - 1)} =$ $= - \frac{2 \ln t}{t^{2} - 1} + \ln (t^{2} - 1) - 2 \ln t =$ $= - \frac{2 t^{2}}{t^{2} - 1} \ln t + \ln (t^{2} - 1) =$ $= - \frac{x + \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}} \ln (x + \sqrt{x^{2} - 1}) + \ln (x^{2} - 1 + x \sqrt{x^{2} - 1}) + C$
	Commented by john santu last updated on 12/May/20

	$thank you sir$
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