Question Number 218427 by mathocean1 last updated on 09/Apr/25
$$ \\ $$$${give}\:{a}\:{recurrence}\:{relation}\:{for}\:{I}_{{n}} . \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }{dx},\:\forall{n}\:\in\:\mathbb{N}. \\ $$
Answered by mehdee7396 last updated on 10/Apr/25
![(1+x^2 )^(−n) =u⇒((−2nx)/((1+x^2 )^(n+1) ))dx dx=dv⇒x=v I_n =x(1+x^2 )^(−n) ]_0 ^1 +2n∫_0 ^1 ((x^2 +1−1)/((1+x^2 )^(n+1) ))dx I_n −2nI_n =2^(−n) −2nI_(n+1) ⇒I_(n+1) =(2^(−n) /(2n))+((2n−1)/(2n))I_n ; I_1 =(π/4)](https://www.tinkutara.com/question/Q218446.png)
$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−{n}} ={u}\Rightarrow\frac{−\mathrm{2}{nx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }{dx} \\ $$$${dx}={dv}\Rightarrow{x}={v} \\ $$$$\left.{I}_{{n}} ={x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−{n}} \right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}{n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }{dx} \\ $$$${I}_{{n}} −\mathrm{2}{nI}_{{n}} =\mathrm{2}^{−{n}} −\mathrm{2}{nI}_{{n}+\mathrm{1}} \\ $$$$\Rightarrow{I}_{{n}+\mathrm{1}} =\frac{\mathrm{2}^{−{n}} }{\mathrm{2}{n}}+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}{I}_{{n}} \:\:\:\:\:\:;\:\:{I}_{\mathrm{1}} =\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$
Commented by mathocean1 last updated on 10/Apr/25
$${Thanks} \\ $$