Question Number 166182 by SANOGO last updated on 14/Feb/22
$${calculer}\:{la}\:{primitive}\:{de} \\ $$$$\int\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$
Answered by greogoury55 last updated on 14/Feb/22
$${t}=\mathrm{tan}\:{u} \\ $$$${Y}=\int\frac{\mathrm{tan}\:^{\mathrm{2}} {u}}{\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {u}\right)^{\mathrm{2}} }\:\mathrm{sec}\:^{\mathrm{2}} {u}\:{du} \\ $$$${Y}=\int\:\frac{\mathrm{tan}\:^{\mathrm{2}} {u}}{\mathrm{sec}\:^{\mathrm{4}} {u}}\:\mathrm{sec}\:^{\mathrm{2}} {u}\:{du} \\ $$$${Y}=\int\:\frac{\mathrm{tan}\:^{\mathrm{2}} {u}}{\mathrm{sec}\:^{\mathrm{2}} {u}}\:{du}=\int\:\mathrm{sin}\:^{\mathrm{2}} {u}\:{du} \\ $$$${Y}=\int\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{u}}{\mathrm{2}}\:{du}\:=\frac{{u}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{u}}{\mathrm{2}}+{c} \\ $$$${Y}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\left({t}\right)−\:\frac{{t}}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}\:+{c}\: \\ $$$$ \\ $$
Commented by SANOGO last updated on 14/Feb/22
$${merci} \\ $$
Answered by MJS_new last updated on 15/Feb/22
![∫(t^2 /((t^2 +1)^2 ))dt= [Ostrogradski′s Method] =−(t/(2(t^2 +1)))+(1/2)∫(dt/(t^2 +1))= =−(t/(2(t^2 +1)))+(1/2)arctan t +C](https://www.tinkutara.com/question/Q166190.png)
$$\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=−\frac{{t}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=−\frac{{t}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:{t}\:+{C} \\ $$