Question Number 68767 by aliesam last updated on 15/Sep/19
Commented by ~ À ® @ 237 ~ last updated on 15/Sep/19
$${Let}\:{named}\:{it}\:\:{f}_{{n}} \left({x}\right) \\ $$$${Let}\:{named}\:\:\forall\:{a}>\mathrm{0}\:\:{g}\left({a},{x}\right)=\int\:\frac{{dx}}{{a}+{x}^{\mathrm{2}} }\: \\ $$$${So}\:\:{g}\left({a},{x}\right)=\:\frac{\mathrm{1}}{{a}}\int\:\:\frac{{dx}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{a}}}\:\int\:\frac{{d}\left(\frac{{x}}{\:\sqrt{{a}}}\:\right)}{\mathrm{1}+\left(\frac{{x}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} }\: \\ $$$${g}\left({a},{x}\right)=\:\frac{{arctan}\left(\frac{{x}}{\:\sqrt{{a}}}\right)}{\:\sqrt{{a}}}\:+{c} \\ $$$${Now}\:{just}\:{ascertain}\:{that}\: \\ $$$$\frac{\partial{g}\left({a},{x}\right)}{\partial{a}}=−\int\:\frac{{dx}}{\left({a}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\frac{\partial^{\mathrm{2}} {g}\left({a},{x}\right)}{\partial{a}^{\mathrm{2}} }=\:\int\:\frac{\mathrm{2}{dx}}{\left({a}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\: \\ $$$$……\:\frac{\partial^{{n}} {g}\left({a},{x}\right)}{\partial{a}^{{n}} }=\:\int\:\frac{\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{1}\right)}{\left({a}+{x}^{\mathrm{2}} \right)^{{n}} }\:{dx} \\ $$$${Finally} \\ $$$$\int\:\frac{{dx}}{\left({a}+{x}^{\mathrm{2}} \right)^{{n}} }\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\mathrm{1}}\:\frac{\partial^{{n}−\mathrm{1}} {g}\left({a},{x}\right)}{\partial{a}^{{n}−\mathrm{1}} } \\ $$$${f}_{{n}} \left({x}\right)=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\mathrm{1}}\:\frac{\partial^{{n}−\mathrm{1}} {g}}{\partial{a}^{{n}−\mathrm{1}} }\left(\mathrm{1},{x}\right) \\ $$