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Question Number 68788 by mhmd last updated on 15/Sep/19

$$\int 1/{(1+x^2)}^{n}dx$$

Commented by Prithwish sen last updated on 15/Sep/19

$$\mathrm{Using}\mathrm{by}\mathrm{parts}$$$${\mathrm{I}}_{\mathrm{n}}=\frac{\mathrm{x}}{{(1+{\mathrm{x}}^2)}^{\mathrm{n}}}+2\mathrm{n}\int \frac{{\mathrm{x}}^2\mathrm{dx}}{{(1+{\mathrm{x}}^2)}^{\mathrm{n}+1}}$$$$=\frac{\mathrm{x}}{{(1+{\mathrm{x}}^2)}^{\mathrm{n}}}+2\mathrm{n}[\int \{\frac{1}{{(1+{\mathrm{x}}^2)}^{\mathrm{n}}}-\frac{1}{{(1+{\mathrm{x}}^2)}^{\mathrm{n}+1}}\}\mathrm{dx}$$$$2{\mathbf{nI}}_{\mathbf{n}+1}-(2\mathbf{n}-1){\mathbf{I}}_{\mathbf{n}}=\frac{\mathrm{x}}{{(1+{\mathrm{x}}^2)}^{\mathrm{n}}}$$$$\mathrm{to}\mathrm{be}\mathrm{continued}.........$$$$$$

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