Question

47) $\int \frac{dx}{(x^2-4)\sqrt{x+1}}$
48) $\int \frac{dx}{(x+1)\sqrt{x^2-1}}$

Answer 1

Dear student
Q 48
$$\begin{gathered} \mathrm{I}=\int \frac{1}{\left(\mathrm{x}+1\right)\sqrt{{\mathrm{x}}^2-1}}\mathrm{dx} \\ \mathrm{Put}\mathrm{x}=\mathrm{secu} \\ \mathrm{So}\mathrm{u}={\sec }^{-1}\mathrm{x} \\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{du}}=\mathrm{secu}\mathrm{tanu} \\ \mathrm{I}=\int \frac{\mathrm{secu}\mathrm{tanudu}}{\left(\mathrm{secu}+1\right)\sqrt{{\sec }^2\mathrm{u}-1}} \\ =\int \frac{\mathrm{secu}}{\mathrm{secu}+1}\mathrm{du} \\ =\int 1\mathrm{du}-\int \frac{1}{\mathrm{secu}+1}\mathrm{du} \\ =\mathrm{u}-\int \frac{1}{{\displaystyle \frac{{\tan }^2{\displaystyle \frac{\mathrm{u}}{2}}+1}{1-{\tan }^2{\displaystyle \frac{\mathrm{u}}{2}}}}+1}\mathrm{du} \\ \mathrm{Put}\mathrm{v}=\tan \frac{\mathrm{u}}{2}\mathrm{so}\frac{\mathrm{dv}}{\mathrm{du}}=\frac{{\sec }^2{\displaystyle \frac{\mathrm{u}}{2}}}{2} \\ \mathrm{I}=\mathrm{u}-\int \frac{{\mathrm{v}}^2-1}{{\mathrm{v}}^2+1}\mathrm{dv} \\ =\mathrm{u}-\left[\int 1\mathrm{dv}-2\int \frac{1}{{\mathrm{v}}^2+1}\mathrm{dv}\right] \\ =\mathrm{u}-\mathrm{v}+2{\tan }^{-1}\mathrm{v}+\mathrm{C} \\ ={\sec }^{-1}\mathrm{x}-\tan \left(\frac{\mathrm{u}}{2}\right)+2{\tan }^{-1}\left(\frac{\mathrm{u}}{2}\right)+\mathrm{C} \\ ={\sec }^{-1}\mathrm{x}-\tan \left(\frac{{\sec }^{-1}\mathrm{x}}{2}\right)+2{\tan }^{-1}\left(\frac{{\sec }^{-1}\mathrm{x}}{2}\right)+\mathrm{C} \end{gathered}$$
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