Question

$\int \frac{1}{\left(x-1\right)\sqrt{x^2-1}}dx$

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\sf \frac{1}{\left(x-1\right)\sqrt{x^2-1}}dx$

To integrate such integrals, we use method of Substitution,

$\rm :\longmapsto\:Put \: x - 1 = \dfrac{1}{y}$

$\rm :\longmapsto\: \: x = \dfrac{1}{y} + 1$

$\rm :\longmapsto\: \: dx = - \dfrac{1}{ {y}^{2} }dy$

On substituting all these values in above integral, we get

$\rm \:  =  \:  \: \displaystyle\int\sf \frac{1}{ \dfrac{1}{y} \sqrt{ {\bigg(\dfrac{1}{y} + 1\bigg) }^{2} - 1} } \times \dfrac{ - 1}{ {y}^{2} } \: dy$

$\rm \:  =  \:  \: \displaystyle\int\sf \frac{1}{ \sqrt{ {\bigg(\dfrac{1}{y} \bigg) }^{2} + 1 + \dfrac{2}{y} - 1} } \times \dfrac{ - 1}{ {y}^{} } \: dy$

$\rm \:  =  \:  \: \displaystyle\int\sf \frac{1}{ \sqrt{ {\bigg(\dfrac{1}{y} \bigg) }^{2} + \dfrac{2}{y}} } \times \dfrac{ - 1}{ {y}^{} } \: dy$

$\rm \:  =  \:  \: \displaystyle\int\sf \frac{1}{\sqrt{ \dfrac{1}{ {y}^{2} } + \dfrac{2}{y}} } \times \dfrac{ - 1}{ {y}^{} } \: dy$

$\rm \:  =  \:  \: \displaystyle\int\sf \frac{1}{\sqrt{ \dfrac{1 + 2y}{ {y}^{2} } } } \times \dfrac{ - 1}{ {y}^{} } \: dy$

$\rm \:  =  \:  \: \displaystyle\int\sf \frac{y}{ \sqrt{1 + 2y} } \times \dfrac{ - 1}{ {y}^{} } \: dy$

$\rm \:  =  \:  - \: \displaystyle\int\sf \frac{1}{ \sqrt{1 + 2y} } \: dy$

$\rm \:  =  \: -   \: \displaystyle\int\sf {\bigg(1 + 2y\bigg) }^{ - \dfrac{1}{2} } \: dy$

$\rm \:  =  \: -   \: \dfrac{ {\bigg(1 + 2y\bigg) }^{ - \dfrac{1}{2} + 1 } }{ - \dfrac{1}{2} + 1} \times \dfrac{1}{2} + c$

$\boxed{ \: \: \: \: \because \: \bf{ \: \displaystyle\int\sf {(ax + b)}^{n} = \frac{ {(ax + b)}^{n + 1} }{a(n + 1)} + c \: \: \: \: \: }}$

$\rm \:  =  \: -   \: \dfrac{ {\bigg(1 + 2y\bigg) }^{\dfrac{1}{2}} }{\dfrac{1}{2}} \times \dfrac{1}{2} + c$

$\rm \:  =  \:  \: - \: \sqrt{1 + 2y} + c$

$\rm \:  =  \:  \: - \: \sqrt{1 + 2 \times \dfrac{1}{x - 1} } + c$

$\rm \:  =  \:  \: - \: \sqrt{1 + \dfrac{2}{x - 1} } + c$

$\rm \:  =  \:  \: - \: \sqrt{\dfrac{x - 1 + 2}{x - 1} } + c$

$\rm \:  =  \:  \: - \: \sqrt{\dfrac{x + 1}{x - 1} } + c$

Additional Information :-

Methods to solve the integral of the form,

$\boxed{ \bf{ \: \: \: \: \: \: \: \displaystyle\int\bf \frac{dx}{ \: \: p \: \: \sqrt{q} \: \: } \: \: \: \: }}$

where,

$\boxed{ \bf{ \: 1. p \: is \: linear, \: q \: is \: linear, \: Substitute \: \sqrt{q} = t}}$

$\boxed{ \bf{ \: 2. p \: is \: quadratic, \: q \: is \: linear, \: Substitute \: \sqrt{q} = t}}$

$\boxed{ \bf{ \: 3. p \: is \: linear, \: q \: is \: quadratic, \: Substitute \: p = \frac{1}{t} }}$

$\boxed{ \bf{ \: 4. p \: is \: quadratic, \: q \: is \: quadratic, \: Substitute \: x = \frac{1}{t} }}$