$Q.17 \int \frac{1 }{(x-1)\sqrt{x2+ 1}}dx =$

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$$\begin{gathered} To Evaluate\hspace{0.17em}: \int \frac{1}{(x-1) \sqrt{x^2+1}}dx \\ Solution : Substituting x= \tan y\Rightarrow y={\tan }^{-1}x \\ \Rightarrow dx= {\sec }^{2}y\mathit{ }dy \\ = \int \frac{1}{(\tan y-1) \sqrt{{\tan }^2 y+1}} {\sec }^{2}y\mathit{ }dy \\ = \int \frac{1}{(\tan y-1) \sqrt{{\sec }^2 y}} {\sec }^{2}y dy \\ = \int \frac{1}{(\tan y-1) \sec y} {\sec }^{2}y dy \\ = \int \frac{\sec y}{\tan y-1} dy \\ = \int \frac{{\displaystyle \frac{1}{\cos y}}}{{\displaystyle \frac{\sin y}{\cos y}}-1} dy \\ = \int \frac{{\displaystyle \frac{1}{\cancel{\cos y}}}}{{\displaystyle \frac{\sin y-\cos y}{\cancel{\cos y}}}} dy \\ = \int \frac{{\displaystyle 1}}{{\displaystyle \sin y-\cos y}} dy \\ Multiplying and dividing the numerator and denominator by \frac{1}{\sqrt{2}} \\ = \int \frac{{\displaystyle \frac{{\displaystyle 1}}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}(\sin y-\cos y)}} dy \\ =\frac{1}{\sqrt{2}} \int \frac{{\displaystyle 1}}{{\displaystyle \frac{1}{\sqrt{2}}\sin y-\frac{1}{\sqrt{2}}\cos y}} dy \\ Since, \frac{1}{\sqrt{2}}= \sin \left(\frac{\mathrm{\pi }}{4}\right) = \cos \left(\frac{\mathrm{\pi }}{4}\right) \\ =\frac{1}{\sqrt{2}} \int \frac{{\displaystyle 1}}{{\displaystyle \sin y \cos \left(\frac{\mathrm{\pi }}{4}\right) - \cos y \sin \left(\frac{\mathrm{\pi }}{4}\right)}} dy \\ Using \sin A \cos B- \cos A \sin B=\sin (A-B) \\ =\frac{1}{\sqrt{2}} \int \frac{{\displaystyle 1}}{{\displaystyle \sin (y- \frac{\mathrm{\pi }}{4})}} dy \\ =\frac{1}{\sqrt{2}} \int \mathrm{cosec} (y- \frac{\mathrm{\pi }}{4}) dy \\ =\frac{1}{\sqrt{2}} \ln \left|\tan \left(\frac{y- \frac{\mathrm{\pi }}{4}}{2}\right)\right|+C \\ =\frac{1}{\sqrt{2}} \ln \left|\tan (\frac{y}{2}-\frac{\mathrm{\pi }}{8})\right|+C \\ substituting y={\tan }^{-1}x \\ =\frac{1}{\sqrt{2}} \ln \left|\tan (\frac{{\tan }^{-1}x}{2}-\frac{\mathrm{\pi }}{8})\right|+C \end{gathered}$$
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