integrate cos^-1((1-x²) / (1+x²)) dx using by parts

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{I}=\int {\cos }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right).\mathrm{dx} \\ \mathrm{We} \mathrm{know} \mathrm{that}: \\ {\cos }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right)=2{\tan }^{-1}\left(\mathrm{x}\right) \\ \Rightarrow \mathrm{I}=2\int {\tan }^{-1}\left(\mathrm{x}\right).\mathrm{dx} \\ =2\int 1.{\tan }^{-1}\left(\mathrm{x}\right).\mathrm{dx} \\ \mathrm{Using} \mathrm{integration} \mathrm{by} \mathrm{part} \mathrm{taking} 1 \mathrm{as} \mathrm{second} \mathrm{function} \mathrm{we} \mathrm{get}: \\ \mathrm{I}=2\left[{\tan }^{-1}\left(\mathrm{x}\right)\int 1.\mathrm{dx}-\int \left\{\frac{\mathrm{d}}{\mathrm{dx}}\left({\tan }^{-1}\left(\mathrm{x}\right)\right)\int 1.\mathrm{dx}\right\}\mathrm{dx}\right] \\ =2\left[{\tan }^{-1}\left(\mathrm{x}\right).\mathrm{x}-\int \left\{\frac{1}{1+{\mathrm{x}}^2}.\mathrm{x}\right\}\mathrm{dx}\right] \\ =2\left[\mathrm{x}.{\tan }^{-1}\left(\mathrm{x}\right)-\int \left\{\frac{\mathrm{x}}{1+{\mathrm{x}}^2}\right\}\mathrm{dx}\right] \\ =2\mathrm{x}.{\tan }^{-1}\left(\mathrm{x}\right)-\int \frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx} \\ \mathrm{Now} \mathrm{in} \int \frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx} \mathrm{if} \mathrm{we} \mathrm{take} \left(1+{\mathrm{x}}^2\right)=\mathrm{t}, \mathrm{then} 2\mathrm{x}.\mathrm{dx}=\mathrm{dt}, \mathrm{hence}, \mathrm{we} \mathrm{get} \\ \int \frac{\mathrm{dt}}{\mathrm{t}}=\ln \left(\mathrm{t}\right)=\ln \left(1+{\mathrm{x}}^2\right) \\ \Rightarrow \mathrm{I}=2\mathrm{x}.{\tan }^{-1}\left(\mathrm{x}\right)-\ln \left(1+{\mathrm{x}}^2\right)+\mathrm{C}, \mathrm{where} \mathrm{C} \mathrm{is} \mathrm{integration} \mathrm{constant}. \\ \Rightarrow \mathbf{\int }{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\right)\mathbf{.}\mathbf{dx}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{.}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{x}\right)\mathbf{-}\mathbf{ln}\left(\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{+}\mathbf{C} \end{gathered}$$

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