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Question-64305




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

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