Question Number 226953 by Spillover last updated on 20/Dec/25
$${If}\:{I}_{{n}} =\int\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}} {dx}\: \\ $$$${Show}\:{that} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{x}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}} +\mathrm{2}{na}^{\mathrm{2}} {I}_{{n}−\mathrm{1}} \: \\ $$
Answered by Spillover last updated on 24/Dec/25
![∫udu=uv−∫vdu u=(x^2 +a^2 )^(n ) du=2nx(x^2 +a^2 ) ∫ dv=dx v=x I_n =x(x^2 +a^2 )^(n ) −∫x.2nx(x^2 +a^2 )^(n−1 ) dx I_n =x(x^2 +a^2 )^(n ) −∫2nx^2 (x^2 +a^2 )^(n−1 ) dx I_n =x(x^2 +a^2 )^(n ) −∫2nx^2 (((x^2 +a^2 )^n )/((x^2 +a^2 )))dx spillover I_n =x(x^2 +a^2 )^(n ) −∫2nx^2 ((((x^2 +a^2 )−a^2 )^n )/((x^2 +a^2 )))dx I_n =x(x^2 +a^2 )^(n ) −2n∫(((x^2 +a^2 )(x^2 +a^2 )^n )/((x^2 +a^2 )))−a^2 ∫(x^2 +a^2 )^(n−1) dx I_n =x(x^2 +a^2 )^(n ) −2n[∫(x^2 +a^2 )^n dx−a^2 ∫(x^2 +a^2 )^(n−1) dx I_n =∫(x^2 +a^2 )^n dx I_(n−1) =∫(x^2 +a^2 )^(n−1) dx I_n =x(x^2 +a^2 )^(n ) −2n[I_n −a^2 I_(n−1) ] I_n (1+2n)=x(x^2 +a^2 )^(n ) +2na^2 I_(n−1) spillover](https://www.tinkutara.com/question/Q226976.png)
$$\int{udu}={uv}−\int{vdu}\: \\ $$$${u}=\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}\:} \:\:{du}=\mathrm{2}{nx}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\:\:\:\:\int\:{dv}={dx}\:\:\:\:\:{v}={x}\: \\ $$$${I}_{{n}} ={x}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}\:} −\int{x}.\mathrm{2}{nx}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}−\mathrm{1}\:} {dx} \\ $$$${I}_{{n}} ={x}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}\:} −\int\mathrm{2}{nx}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}−\mathrm{1}\:} {dx} \\ $$$${I}_{{n}} ={x}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}\:} −\int\mathrm{2}{nx}^{\mathrm{2}} \frac{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}} }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{dx}\:\:\:\:\:\:\:\:\:\:\:{spillover} \\ $$$${I}_{{n}} ={x}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}\:} −\int\mathrm{2}{nx}^{\mathrm{2}} \frac{\left(\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} \right)^{{n}} }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{dx} \\ $$$${I}_{{n}} ={x}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}\:} −\mathrm{2}{n}\int\frac{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}} }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}−{a}^{\mathrm{2}} \int\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} {dx} \\ $$$${I}_{{n}} ={x}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}\:} −\mathrm{2}{n}\left[\int\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}} {dx}−{a}^{\mathrm{2}} \int\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} {dx}\right. \\ $$$${I}_{{n}} =\int\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}} {dx}\:\:\:\:\:\:{I}_{{n}−\mathrm{1}} =\int\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} {dx} \\ $$$${I}_{{n}} ={x}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}\:} −\mathrm{2}{n}\left[{I}_{{n}} −{a}^{\mathrm{2}} {I}_{{n}−\mathrm{1}} \right] \\ $$$${I}_{{n}} \left(\mathrm{1}+\mathrm{2}{n}\right)={x}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{{n}\:} +\mathrm{2}{na}^{\mathrm{2}} {I}_{{n}−\mathrm{1}} \:\:\:\:\:\:\:\:{spillover} \\ $$$$ \\ $$$$ \\ $$