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Question-68767




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 (x)  Let named  ∀ a>0  g(a,x)=∫ (dx/(a+x^2 ))   So  g(a,x)= (1/a)∫  (dx/(1+(x^2 /a))) = (1/( (√a))) ∫ ((d((x/( (√a))) ))/(1+((x/( (√a))))^2 ))   g(a,x)= ((arctan((x/( (√a)))))/( (√a))) +c  Now just ascertain that   ((∂g(a,x))/∂a)=−∫ (dx/((a+x^2 )^2 ))  ((∂^2 g(a,x))/∂a^2 )= ∫ ((2dx)/((a+x^2 )^3 ))   ...... ((∂^n g(a,x))/∂a^n )= ∫ (((−1)^n (n−1))/((a+x^2 )^n )) dx  Finally  ∫ (dx/((a+x^2 )^n )) = (((−1)^n )/(n−1)) ((∂^(n−1) g(a,x))/∂a^(n−1) )  f_n (x)= (((−1)^n )/(n−1)) (∂^(n−1) g/∂a^(n−1) )(1,x)
$${Let}\:{named}\:{it}\:\:{f}_{{n}} \left({x}\right) \\ $$$${Let}\:{named}\:\:\forall\:{a}>\mathrm{0}\:\:{g}\left({a},{x}\right)=\int\:\frac{{dx}}{{a}+{x}^{\mathrm{2}} }\: \\ $$$${So}\:\:{g}\left({a},{x}\right)=\:\frac{\mathrm{1}}{{a}}\int\:\:\frac{{dx}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{a}}}\:\int\:\frac{{d}\left(\frac{{x}}{\:\sqrt{{a}}}\:\right)}{\mathrm{1}+\left(\frac{{x}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} }\: \\ $$$${g}\left({a},{x}\right)=\:\frac{{arctan}\left(\frac{{x}}{\:\sqrt{{a}}}\right)}{\:\sqrt{{a}}}\:+{c} \\ $$$${Now}\:{just}\:{ascertain}\:{that}\: \\ $$$$\frac{\partial{g}\left({a},{x}\right)}{\partial{a}}=−\int\:\frac{{dx}}{\left({a}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\frac{\partial^{\mathrm{2}} {g}\left({a},{x}\right)}{\partial{a}^{\mathrm{2}} }=\:\int\:\frac{\mathrm{2}{dx}}{\left({a}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\: \\ $$$$……\:\frac{\partial^{{n}} {g}\left({a},{x}\right)}{\partial{a}^{{n}} }=\:\int\:\frac{\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{1}\right)}{\left({a}+{x}^{\mathrm{2}} \right)^{{n}} }\:{dx} \\ $$$${Finally} \\ $$$$\int\:\frac{{dx}}{\left({a}+{x}^{\mathrm{2}} \right)^{{n}} }\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\mathrm{1}}\:\frac{\partial^{{n}−\mathrm{1}} {g}\left({a},{x}\right)}{\partial{a}^{{n}−\mathrm{1}} } \\ $$$${f}_{{n}} \left({x}\right)=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\mathrm{1}}\:\frac{\partial^{{n}−\mathrm{1}} {g}}{\partial{a}^{{n}−\mathrm{1}} }\left(\mathrm{1},{x}\right) \\ $$

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