Question Number 47101 by Meritguide1234 last updated on 04/Nov/18 | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18 | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18 | ||
$$\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}×\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}−\mathrm{1}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} } \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\mid\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} }\mid_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{I}_{{n}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}×\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{I}_{{n}−\mathrm{1}} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)}×\frac{\mathrm{1}}{\mathrm{2}^{{n}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{I}_{{n}−\mathrm{1}} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{1}} }=\frac{\pi}{\mathrm{4}} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)}×\frac{\mathrm{1}}{\mathrm{2}^{{n}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{I}_{{n}−\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\pi}{\mathrm{8}} \\ $$$${I}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\pi}{\mathrm{8}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{3}\pi}{\mathrm{32}} \\ $$$${contd}... \\ $$$$ \\ $$ | ||