Question Number 190740 by mathlove last updated on 10/Apr/23
$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\mathrm{2}} }=? \\ $$
Answered by gatocomcirrose last updated on 10/Apr/23
$$\mathrm{x}=\sqrt{\mathrm{5}}\mathrm{tg}\theta\Rightarrow\mathrm{dx}=\sqrt{\mathrm{5}}\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\int\frac{\sqrt{\mathrm{5}}\mathrm{sec}^{\mathrm{2}} \theta}{\mathrm{25sec}^{\mathrm{4}} \theta}\mathrm{d}\theta=\frac{\sqrt{\mathrm{5}}}{\mathrm{25}}\int\mathrm{cos}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{50}}\int\mathrm{cos}\left(\mathrm{2}\theta\right)+\mathrm{1d}\theta=\frac{\sqrt{\mathrm{5}}}{\mathrm{100}}\mathrm{sin}\left(\mathrm{2}\theta\right)+\theta \\ $$$$=\frac{\mathrm{x}}{\mathrm{10}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{5}\right)}+\mathrm{arctan}\left(\frac{\mathrm{x}}{\:\sqrt{\mathrm{5}}}\right)+\mathrm{C} \\ $$
Commented by mehdee42 last updated on 10/Apr/23
$${very}\:{good} \\ $$$${decreasing}\:{order} \\ $$$$\:{if}\:\:\:{I}_{{n}} =\int\:\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}\right)^{{n}} }\Rightarrow{I}_{{n}+\mathrm{1}} \:=\:\frac{{x}}{\mathrm{2}{na}\left({x}^{\mathrm{2}} +{a}\right)^{{n}} }+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{na}}\:{I}_{{n}} \\ $$