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Question Number 64305 by aliesam last updated on 16/Jul/19

Commented by aliesam last updated on 16/Jul/19

$p r o v e t h a t$
Commented by mathmax by abdo last updated on 16/Jul/19

$l e t J_{n} = \int \frac{d t}{{(1 + t^{2})}^{n}} \Rightarrow J_{n} = \int \frac{1 + t^{2}}{{(1 + t^{2})}^{n + 1}} d t$ $= \int \frac{d t}{{(1 + t^{2})}^{n + 1}} + \int \frac{t^{2}}{{(1 + t^{2})}^{n + 1}} d t$ $\int \frac{d t}{{(1 + t^{2})}^{n + 1}} = J_{n + 1} a n d b y p a r t s u = t a n t v^{^{'}} = t {(1 + t^{2})}^{- n - 1}$ $\int \frac{t^{2}}{{(1 + t^{2})}^{n + 1}} d t = t (- \frac{1}{2 n} {(1 + t^{2})}^{- n}) - \int - \frac{1}{2 n} {(1 + t^{2})}^{- n} d t$ $= - \frac{1}{2 n} \frac{t}{{(1 + t^{2})}^{n}} + \frac{1}{2 n} J_{n} \Rightarrow J_{n} = J_{n + 1} - \frac{1}{2 n} \frac{t}{{(1 + t^{2})}^{n}} + \frac{1}{2 n} J_{n} \Rightarrow$ $(1 - \frac{1}{2 n}) J_{n} = - \frac{t}{2 n {(1 + t^{2})}^{n}} + J_{n + 1} \Rightarrow$ $(2 n - 1) J_{n} = 2 n J_{n + 1} - \frac{t}{{(1 + t^{2})}^{n}}$
Commented by mathmax by abdo last updated on 16/Jul/19

$\Rightarrow 2 n J_{n + 1} = (2 n - 1) J_{n} + \frac{t}{{(1 + t^{2})}^{n}} \Rightarrow$ $J_{n + 1} = \frac{2 n - 1}{2 n} J_{n} + \frac{t}{2 n {(1 + t^{2})}^{n}}$
Commented by aliesam last updated on 16/Jul/19

$t h a n k y o u s i r$ $b u t w h a t a b o u t s t a r t i n g f r o m j_{n} + 1 ?$
Commented by mathmax by abdo last updated on 16/Jul/19

$i n o r d r t o a p p e a r J_{n}$
Commented by aliesam last updated on 16/Jul/19

$y e s s i r$
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