Question Number 28703 by students last updated on 29/Jan/18
$$\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:\:\:{solve}\:{the}\:{integration} \\ $$
Commented by abdo imad last updated on 29/Jan/18
$${let}\:{put}\:{I}=\:\int\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{if}\:{a}\neq\mathrm{0}\:\:{the}\:{ch}.\:{x}\:={atan}\theta\:{give} \\ $$$${I}=\:\:\int\:\:\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} \left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right.}\:{a}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right) \\ $$$$=\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:\int\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}} {d}\theta\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:\:\int\:\:\frac{{d}\theta}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}} \\ $$$$=\frac{\xi}{{a}^{\mathrm{2}} }\:\int\:{cos}\theta\:{d}\theta\:=\frac{\xi}{{a}^{\mathrm{2}} }\:{sin}\theta\:+{k}\:=\frac{\xi}{{a}^{\mathrm{2}} }\:{sin}\left({arctan}\left(\frac{{x}}{{a}}\right)\right)\:+{k} \\ $$$${and}\:\:\xi^{\mathrm{2}} =\mathrm{1}. \\ $$$${if}\:{a}=\mathrm{0}\:\:{I}=\:\int\:{x}^{−\mathrm{3}} {dx}\:=\:\frac{\mathrm{1}}{−\mathrm{3}+\mathrm{1}}\:{x}^{−\mathrm{3}+\mathrm{1}} \:+{k} \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+{k}. \\ $$$$ \\ $$$$ \\ $$