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

Dear Student,

$$\begin{gathered} \frac{\sqrt{x + 1} - \sqrt{x - 1}}{\sqrt{x + 1} + \sqrt{x - 1}} \\ = \frac{{\displaystyle \sqrt{x + 1} - \sqrt{x - 1}}}{{\displaystyle \sqrt{x + 1} + \sqrt{x - 1}}} \times \frac{{\displaystyle \sqrt{x + 1} - \sqrt{x - 1}}}{{\displaystyle \sqrt{x + 1} - \sqrt{x - 1}}} \left[Rationali\sin g the denominator\right] \\ = \frac{{\displaystyle {\left(\sqrt{x + 1} - \sqrt{x - 1}\right)}^2}}{{\displaystyle {\left(\sqrt{x + 1}\right)}^2 - {\left(\sqrt{x - 1}\right)}^2}} \\ = \frac{x + \cancel{1} + x - \cancel{1} + 2\sqrt{\left(x + 1\right)\left(x - 1\right)}}{\cancel{x} + 1 - \cancel{x} + 1} \\ = \frac{{\displaystyle 2x - 2\sqrt{x^2 - 1}}}{2} \\ = x - \sqrt{x^2 - 1} \\ \therefore \int \frac{{\displaystyle \sqrt{x + 1} - \sqrt{x - 1}}}{{\displaystyle \sqrt{x + 1} + \sqrt{x - 1}}}dx \\ = \int \left(x - \sqrt{x^2 - 1}\right)dx \\ = \int xdx - \int \sqrt{x^2 - 1^2}dx \\ = \frac{x^2}{2} - \left[\frac{x}{2}\sqrt{x^2 - 1} - \frac{1}{2}\log \left|x + \sqrt{x^2 - 1}\right|\right] + C \\ = \boxed{\frac{{\displaystyle {\mathbf{x}}^{\mathbf{2}}}}{{\displaystyle \mathbf{2}}}\mathbf{ }\mathbf{-}\mathbf{ }\frac{{\displaystyle \mathbf{x}}}{{\displaystyle \mathbf{2}}}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{1}}\mathbf{ }\mathbf{+}\mathbf{ }\frac{{\displaystyle \mathbf{1}}}{{\displaystyle \mathbf{2}}}\mathbf{log}\left|\mathbf{x}\mathbf{ }\mathbf{+}\mathbf{ }\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{1}}\right|\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{C}} \end{gathered}$$

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