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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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