Question

integrate :

1/[(x^2-2)sqrt(x^2-1)] dx

Answer 1

I think this might be able to help. c; http://prntscr.com/apfe8g

Answer 2

this doesn't give the same answer as Wolfram but I think it looks ok-ish

\(I = \int ~ \dfrac{1}{(x^2 - 2) \sqrt {x^2 -1}} ~ dx\)

\(x = \cosh u, ~ dx = \sinh u ~ du\)

so


\(\int \dfrac{1}{(\cosh^2 u - 2) \sqrt {\cosh^2 u -1}} ~ \sinh u ~ du \)

\(= \int \dfrac{1}{(\cosh^2 u - 2) \sinh u} ~ \sinh u ~ du \)

\(= \int \dfrac{1}{(\sinh^2 u - 1) } ~ du \)


then \(\sec p = \sinh u, ~ \sec p \tan p ~ dp = \cosh u ~ du, ~ du = \dfrac{\sec p \tan p }{\cosh u}~ dp\)

and

\(\cosh u = \sqrt{\sec^2 p + 1}\)


\(\Rightarrow \int \dfrac{1}{(\sec^2 p - 1) } ~ \dfrac{\sec p \tan p}{\sqrt{\sec^2 p + 1}} ~ dp \)

\(\Rightarrow \int \dfrac{1}{\tan^2 p } ~ \dfrac{\sec p \tan p}{\sqrt{\sec^2 p + 1}} ~ dp \)



\(\Rightarrow \int \dfrac{1}{\frac{\sin p}{\cos p} } ~ \dfrac{\frac{1}{\cos p} }{\sqrt{(\frac{1}{\cos p})^2 + 1}} ~ dp \)



\(\Rightarrow \int ~ \dfrac{1 }{\sqrt{\tan^2 p + 1}} ~ dp \)



\(\Rightarrow \int ~ \cos p ~ dp \)

\( = \sin p + C\) !!!

from \(\sec p = \sinh u\)


\(\sin p = \sqrt{1 - cosech^2 u} \)


So, \(I = \sqrt{1 - cosech^2 \cosh^{-1} x} ~ + C \)

Answer 3

@Ms-Brains , Not sure how to take it from there.

I tried to start it by letting

u = x^2 - 1

ending up at a stalemate.

@IrishBoy123, I think this step should be ? (excuse me, If I am mistaken)

\[\int\limits_{}^{}\frac{ 1 }{ \sin p \sqrt{(\frac{ 1 }{ \cos^2x}) + 1}}\]

\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{\tan^2x + \sin^2p}}\]

Answer 4

* replace the x with p

Answer 5

you're quite right! thank you.

Answer 6

\[u^2 = x^2 - 1\]

\[dx = \frac{ u du }{ \sqrt{u^2 +1} }\]

\[\frac{ \frac{ udu }{ \sqrt{u^2 + 1}} }{ (u^2 - 1) u}\]

\[\int\limits_{}^{}\frac{ du }{ (u^2 - 1)\sqrt{u^2 + 1} }\]

let u = tantheta

\[u = \tan \theta , du = \sec^2 \theta\]

\[\int\limits_{}^{} \frac{ \sec^2 \theta d \theta }{ (\tan^2 \theta - 1) \sec \theta }\]

\[\int\limits_{}^{}\frac{ \frac{ 1 }{ \cos \theta } d \theta }{ \frac{ \sin^2 \theta }{ \cos ^2 \theta } - 1 }\]


\[\int\limits_{}^{}\frac{ \cos \theta d \theta }{ -(\cos^2 \theta - \sin^2 \theta) } = \int\limits_{}^{} \frac{ d \sin \theta }{ 2 \sin^2 \theta - 1 }\]

\[\frac{ 1 }{ \sqrt(2) } \int\limits_{}^{} \frac{ d[\sqrt{2} \sin \theta] }{ [\sqrt{2}\sin \theta]^2 - 1}\]

\[\frac{ -1 }{ \sqrt{2} } \tanh^-1(\sqrt{2}\sin \theta) + c\]

\[\frac{ -1 }{ \sqrt{2} } \tanh^{-1}(\sqrt(2) \frac{ \sqrt{x^2 - 1} }{ x }) + c\]

Answer 7

cool!