Question

\[\int\frac{1}{\left( x - 1 \right) \sqrt{x^2 + 1}} \text{ dx }\]

Answer 1

\[\text{ We  have, } \]
\[I = \int \frac{dx}{\left( x - 1 \right) \sqrt{x^2 + 1}}\]
\[\text{ Putting  x }- 1 = \frac{1}{t}\]
\[ \Rightarrow dx = - \frac{1}{t^2}dt\]
\[ \therefore I = \int\frac{- \frac{1}{t^2}dt}{\left( \frac{1}{t} \right) \sqrt{\left( 1 + \frac{1}{t} \right)^2 + 1}}\]
\[ = \int \frac{- \frac{1}{t}dt}{\sqrt{1 + \frac{1}{t^2} + \frac{2}{t} + 1}}\]
\[ = \int \frac{- \frac{1}{t}dt}{\frac{\sqrt{t^2 + 1 + 2t + t^2}}{t}}\]
\[ = \int \frac{- dt}{\sqrt{2 t^2 + 2t + 1}}\]
\[ = - \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{t^2 + t + \frac{1}{2}}}\]
\[ = - \frac{1}{\sqrt{2}}\int \frac{dt}{\sqrt{t^2 + t + \frac{1}{4} - \frac{1}{4} + \frac{1}{2}}}\]
\[ = - \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{\left( t + \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2}}\]
\[ = - \frac{1}{\sqrt{2}}\text{ log }\left| t + \frac{1}{2} + \sqrt{\left( t + \frac{1}{2} \right)^2 + \frac{1}{4}} \right| + \text{ C where t} = \frac{1}{x - 1}\]