Question Number 35682 by prof Abdo imad last updated on 22/May/18
$${let}\:{F}\left({x}\right)\:=\:\int_{{x}\:+\mathrm{1}} ^{{x}^{\mathrm{2}} \:+\mathrm{1}} \:\:\:{arctan}\left(\mathrm{1}+{t}\right){dt} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\frac{\partial{F}}{\partial{x}}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:{F}\left({x}\right)\:. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/May/18
Commented by abdo mathsup 649 cc last updated on 23/May/18
![1) we have (∂F/∂x)(x) =(x^2 +1)^′ arctan(2+x^2 ) −(x+1)^′ arctan( 2+x) =2x arctan(3+x^2 ) −arctan(2+x) 2) ∃ ξ ∈]x+1,x^2 +1[ / F(x) = arctan(1+ξ) ∫_(x+1) ^(x^2 +1) dt =(x^2 −x) arctan(1+ξ) ⇒ lim_(x→0) F(x) = 0×arctan(2)=0 .](https://www.tinkutara.com/question/Q35767.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\frac{\partial{F}}{\partial{x}}\left({x}\right)\:=\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{'} \:{arctan}\left(\mathrm{2}+{x}^{\mathrm{2}} \right) \\ $$$$−\left({x}+\mathrm{1}\right)^{'} \:{arctan}\left(\:\mathrm{2}+{x}\right) \\ $$$$=\mathrm{2}{x}\:{arctan}\left(\mathrm{3}+{x}^{\mathrm{2}} \right)\:−{arctan}\left(\mathrm{2}+{x}\right) \\ $$$$\left.\mathrm{2}\left.\right)\:\exists\:\xi\:\in\right]{x}+\mathrm{1},{x}^{\mathrm{2}} +\mathrm{1}\left[\:\:/\right. \\ $$$${F}\left({x}\right)\:=\:{arctan}\left(\mathrm{1}+\xi\right)\:\int_{{x}+\mathrm{1}} ^{{x}^{\mathrm{2}} \:+\mathrm{1}} {dt} \\ $$$$=\left({x}^{\mathrm{2}} −{x}\right)\:{arctan}\left(\mathrm{1}+\xi\right)\:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}} {F}\left({x}\right) \\ $$$$=\:\mathrm{0}×{arctan}\left(\mathrm{2}\right)=\mathrm{0}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18
$$\int{tan}^{−\mathrm{1}} \left(\mathrm{1}+{t}\right){dt} \\ $$$$={ttan}^{−\mathrm{1}} \left(\mathrm{1}+{t}\right)−\int\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }×{tdt} \\ $$$$={t}×{tan}^{−\mathrm{1}} \left({t}+\mathrm{1}\right)−\int\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}}{dt} \\ $$$$={t}×{tan}^{−\mathrm{1}} \left({t}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{t}+\mathrm{2}−\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}}{dt} \\ $$$$={t}×{tan}^{−\mathrm{1}} \left({t}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{t}+\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}}+\int\frac{{dt}}{\mathrm{1}+\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$={t}×{tan}^{−\mathrm{1}} \left({t}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}×{ln}\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)+{tan}^{−\mathrm{1}} \left({t}+\mathrm{1}\right. \\ $$$${now}\:{putting}\:{the}\:{upper}\:{limit}\:{and}\:{lower}\:{limit} \\ $$$${up}=\left({x}^{\mathrm{2}} +\mathrm{1}\right){tan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}×{ln}\left({x}^{\mathrm{4}} +\mathrm{2}^{} {x}^{\mathrm{2}} +\mathrm{1}+\right. \\ $$$$\left.\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}+\mathrm{2}\right)+{tan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{2}\right) \\ $$$$ \\ $$$${lw}=\left({x}+\mathrm{1}\right)×{tan}^{−\mathrm{1}} \left({x}+\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}×{ln}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}+\right. \\ $$$$\left.\mathrm{2}{x}+\mathrm{2}+\mathrm{2}\right)+{tan}^{−\mathrm{1}} \left({x}+\mathrm{2}\right) \\ $$$${required}\:{result}\:=\frac{\partial}{\partial{x}}\left({up}−{lw}\right)\:\: \\ $$$${page}\:{area}\:{not}\:{sufficient}\:… \\ $$