Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given integral is

[tex]\displaystyle{\bf{\;\;{\int \frac{ dx}{(x - 1) \sqrt{( {x}^{2} + 1)} }}}} \\ [/tex]

To solve this integral, we use method of Substitution.

So, Substitute

[tex]\rm \: x - 1 = \dfrac{1}{y} \\ [/tex]

[tex]\rm \: x = \dfrac{1}{y} + 1 \\ [/tex]

[tex]\rm \: dx = \: - \: \dfrac{1}{ {y}^{2} } \: dy \\ [/tex]

So, on substituting these values in above integral, we get

[tex]\rm \: = \: - \: \displaystyle\int\rm \frac{1}{\dfrac{1}{y} \sqrt{ {\bigg(\dfrac{1}{y} + 1 \bigg)}^{2} + 1} } \times \frac{1}{ {y}^{2}} \: dy \\ [/tex]

[tex]\rm \: = \: - \: \displaystyle\int\rm \frac{1}{ \sqrt{ {\bigg(\dfrac{1}{y} + 1 \bigg)}^{2} + 1} } \times \frac{1}{ {y}} \: dy \\ [/tex]

[tex]\rm \: = \: - \: \displaystyle\int\rm \frac{1}{ \sqrt{ {\bigg(\dfrac{1 + y}{y}\bigg)}^{2} + 1}} \times \frac{1}{ {y}} \: dy \\ [/tex]

[tex]\rm \: = \: - \: \displaystyle\int\rm \frac{1}{ \sqrt{\dfrac{ {y}^{2} + 1 + 2y }{ {y}^{2} } + 1} } \times \frac{1}{ {y}} \: dy \\ [/tex]

[tex]\rm \: = \: - \: \displaystyle\int\rm \frac{1}{ \sqrt{\dfrac{ {y}^{2} + 1 + 2y + {y}^{2} }{ {y}^{2} }} } \times \frac{1}{ {y}} \: dy \\ [/tex]

[tex]\rm \: = \: - \: \displaystyle\int\rm \frac{y}{ \sqrt{ {2y}^{2} + 2y + 1 } } \times \frac{1}{y} \: dy \\ [/tex]

[tex]\rm \: = \: - \: \displaystyle\int\rm \frac{1}{ \sqrt{ {2y}^{2} + 2y + 1 } } \: dy \\ [/tex]

[tex]\rm \: = \: - \: \dfrac{1}{ \sqrt{2} } \displaystyle\int\rm \frac{1}{ \sqrt{ {y}^{2} + y + \dfrac{1}{2} } } \: dy \\ [/tex]

[tex]\rm \: = \: - \: \dfrac{1}{ \sqrt{2} } \displaystyle\int\rm \frac{1}{ \sqrt{ {y}^{2} + y + \dfrac{1}{4} - \dfrac{1}{4} + \dfrac{1}{2} } } \: dy \\ [/tex]

[tex]\rm \: = \: - \: \dfrac{1}{ \sqrt{2} } \displaystyle\int\rm \frac{1}{ \sqrt{ {\bigg(y + \dfrac{1}{2} \bigg) }^{2} + \dfrac{1}{4}} } \: dy \\ [/tex]

[tex]\rm \: = \: - \: \dfrac{1}{ \sqrt{2} } \displaystyle\int\rm \frac{1}{ \sqrt{ {\bigg(y + \dfrac{1}{2} \bigg) }^{2} + {\bigg(\dfrac{1}{2} \bigg) }^{2} } } \: dy \\ [/tex]

We know,

[tex]\boxed{ \rm{ \:\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log |x + \sqrt{ {x}^{2} + {a}^{2}} | + c \: }} \\ [/tex]

So, using this, we get

[tex]\rm \: = \: - \: \dfrac{1}{ \sqrt{2} } log\bigg |y + \dfrac{1}{2} + \sqrt{ {\bigg(y + \dfrac{1}{2}\bigg) }^{2} + {\bigg(\dfrac{1}{2} \bigg) }^{2} } \bigg| + c\\ [/tex]

[tex]\rm \: = \: - \: \dfrac{1}{ \sqrt{2} } log\bigg |y + \dfrac{1}{2} + \sqrt{ {y}^{2} + y + \dfrac{1}{2} } \bigg| + c\\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log |x + \sqrt{ {x}^{2} - {a}^{2} } | + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log |x + \sqrt{ {x}^{2} + {a}^{2}} | + c}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]