Question Number 64305 by aliesam last updated on 16/Jul/19
Commented by aliesam last updated on 16/Jul/19
$${prove}\:{that} \\ $$
Commented by mathmax by abdo last updated on 16/Jul/19
$${let}\:{J}_{{n}} =\int\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }\:\Rightarrow{J}_{{n}} =\int\:\:\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }{dt} \\ $$$$=\int\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\:+\int\:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\:{dt}\: \\ $$$$\int\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\:={J}_{{n}+\mathrm{1}} \:\:\:\:\:{and}\:{by}\:{parts}\:{u}\:={t}\:\:{ant}\:{v}^{'} \:={t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−{n}−\mathrm{1}} \\ $$$$\int\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }{dt}\:={t}\:\left(−\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{n}}}\left(\mathrm{1}+\boldsymbol{{t}}^{\mathrm{2}} \right)^{−\boldsymbol{{n}}} \right)−\int\:\:\:−\frac{\mathrm{1}}{\mathrm{2}{n}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−{n}} {dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{n}}\:\:\frac{{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }\:\:+\frac{\mathrm{1}}{\mathrm{2}{n}}\:{J}_{{n}} \:\Rightarrow{J}_{{n}} ={J}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}{n}}\frac{{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }\:+\frac{\mathrm{1}}{\mathrm{2}{n}}\:{J}_{{n}} \:\Rightarrow \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right){J}_{{n}} =−\frac{{t}}{\mathrm{2}{n}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }\:+{J}_{{n}+\mathrm{1}} \:\Rightarrow \\ $$$$\left(\mathrm{2}{n}−\mathrm{1}\right){J}_{{n}} =\:\mathrm{2}{n}\:{J}_{{n}+\mathrm{1}} −\frac{{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} } \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 16/Jul/19
$$\Rightarrow\mathrm{2}{n}\:{J}_{{n}+\mathrm{1}} =\left(\mathrm{2}{n}−\mathrm{1}\right){J}_{{n}} \:\:+\frac{{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }\:\Rightarrow \\ $$$${J}_{{n}+\mathrm{1}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}\:{J}_{{n}} \:\:+\frac{{t}}{\mathrm{2}{n}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} } \\ $$
Commented by aliesam last updated on 16/Jul/19
$${thank}\:{you}\:{sir}\: \\ $$$${but}\:{what}\:{about}\:{starting}\:{from}\:{j}_{{n}} +\mathrm{1}\:? \\ $$
Commented by mathmax by abdo last updated on 16/Jul/19
$${in}\:{ordr}\:{to}\:{appear}\:{J}_{{n}} \\ $$
Commented by aliesam last updated on 16/Jul/19
$${yes}\:{sir} \\ $$