Question Number 41135 by math khazana by abdo last updated on 02/Aug/18
$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({xt}\right){dt}\:\:{x}\:{from}\:{R}\: \\ $$
Answered by math khazana by abdo last updated on 03/Aug/18
![we have f^′ (x)= ∫_0 ^1 (t/(1+x^2 t^2 ))dt =_(xt =u) ∫_0 ^x (u/(x(1+u^2 ))) (du/x) = (1/x^2 ) ∫_0 ^x ((udu)/(1+u^2 )) =(1/(2x^2 ))[ln(1+u^2 )]_0 ^x =((ln(1+x^2 ))/(2x^2 )) ⇒ f(x) = ∫_0 ^x ((ln(1+t^2 ))/(2t^2 )) dt +c c=f(0) =0 ⇒ f(x)=∫_0 ^x ((ln(1+t^2 ))/(2t^2 )) dt by parts 2f(x)=[−(1/t)ln(1+t^2 )]_0 ^x + ∫_0 ^x (1/t) ((2t)/(1+t^2 ))dt =−(1/x)ln(1+x^2 ) +2arctanx ⇒ f(x)= arctan(x)−(1/(2x))ln(1+x^2 ) .](Q41184.png)
$${we}\:{have}\:{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }{dt} \\ $$$$=_{{xt}\:={u}} \:\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{{u}}{{x}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\frac{{du}}{{x}}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{udu}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\left[{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{{x}} \:\:=\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{2}{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}^{\mathrm{2}} }\:{dt}\:+{c} \\ $$$${c}={f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}^{\mathrm{2}} }\:{dt} \\ $$$${by}\:{parts}\:\mathrm{2}{f}\left({x}\right)=\left[−\frac{\mathrm{1}}{{t}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{{x}} \:+\:\int_{\mathrm{0}} ^{{x}} \:\frac{\mathrm{1}}{{t}}\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=−\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\:+\mathrm{2}{arctanx}\:\:\Rightarrow \\ $$$${f}\left({x}\right)=\:{arctan}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}{x}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:. \\ $$
Commented by math khazana by abdo last updated on 03/Aug/18
$${let}\:{prove}\:{lim}_{{t}\rightarrow\mathrm{0}} \:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{{t}}\:=\mathrm{0}\:{we}\:{have} \\ $$$${ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\sim\:{t}^{\mathrm{2}} \Rightarrow\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{{t}}\sim\:{t}\:\left({t}\rightarrow\mathrm{0}\right)\:{result}\:{is}\:{proved} \\ $$