Question Number 62437 by mathsolverby Abdo last updated on 21/Jun/19
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left(\mathrm{1}+{xt}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$${determine}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left(\mathrm{1}+\mathrm{2}{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 28/Jun/19
![1) we have f^′ (x) =∫_0 ^1 (t/((1+x^2 t^2 )(t^2 +1)))dt =_(xt =u) ∫_0 ^x (u/(x(1+u^2 )(1+(u^2 /x^2 )))) (du/x) =∫_0 ^x ((udu)/((1+u^2 )(x^2 +u^2 ))) let decompose F(u) =(u/((u^2 +1)(u^2 +x^2 ))) ⇒ F(u) =((au +b)/(u^2 +1)) +((cu +d)/(u^2 +x^2 )) F(−u) =−F(u)⇒((−au +b)/(u^2 +1)) +((−cu +d)/(u^2 +x^2 )) =((−au−b)/(u^2 +1)) +((−cu−d)/(u^2 +x^2 )) ⇒b=d =0 ⇒ F(u) =((au)/(u^2 +1)) +((cu)/(u^2 +x^2 )) lim_(u→+∞) uF(u) =0 =a+c ⇒c=−a ⇒F(u) =((au)/(u^2 +1)) −((au)/(u^2 +x^2 )) F(1) =(1/(2(x^2 +1))) =(a/2) −(a/(x^2 +1)) ⇒((ax^2 +a−2a)/(2(x^(2 ) +1))) =(1/(2(x^2 +1))) ⇒ax^2 −a =1 ⇒ a(x^2 −1) =1 ⇒ a =(1/(x^2 −1)) (we suppose x≠+^− 1 and x≠0) ⇒ F(u) =(1/(x^2 −1)){(u/(u^2 +1)) −(u/(u^2 +x^2 ))}⇒f^′ (x) =∫_0 ^x F(u)du =(1/(x^2 −1)) { ∫_0 ^x ((udu)/(u^2 +1)) −∫_0 ^x ((udu)/(u^2 +x^2 ))} we have ∫_0 ^x ((udu)/(u^2 +1)) =[(1/2)ln(u^2 +1)]_0 ^x =(1/2)ln(x^2 +1) ∫_0 ^x ((udu)/(u^2 +x^2 )) =_(u =xα) ∫_0 ^1 ((xα xdα)/(x^2 (1+α^2 ))) =∫_0 ^1 ((αdα)/(1+α^2 )) =[(1/2)ln(1+α^2 )]_0 ^1 =((ln(2))/2) ⇒ f^′ (x) =((ln(x^2 +1))/(2(x^2 −1))) −((ln(2))/(2(x^2 −1))) ⇒f(x) =∫ ((ln(1+x^2 ))/(2(x^2 −1)))dx −((ln(2))/2) ∫ (dx/(x^2 −1)) +c .....be continued....](https://www.tinkutara.com/question/Q63027.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}}{\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt}\:=_{{xt}\:={u}} \:\:\:\:\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{u}}{{x}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}\:\frac{{du}}{{x}} \\ $$$$=\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{{udu}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{u}^{\mathrm{2}} \right)}\:\:{let}\:{decompose}\:{F}\left({u}\right)\:=\frac{{u}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{au}\:+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} } \\ $$$${F}\left(−{u}\right)\:=−{F}\left({u}\right)\Rightarrow\frac{−{au}\:+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{cu}\:+{d}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:=\frac{−{au}−{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{cu}−{d}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:\Rightarrow{b}={d}\:=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{au}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{cu}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} } \\ $$$${lim}_{{u}\rightarrow+\infty} {uF}\left({u}\right)\:=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow{F}\left({u}\right)\:=\frac{{au}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{au}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{a}}{\mathrm{2}}\:−\frac{{a}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\frac{{ax}^{\mathrm{2}} \:+{a}−\mathrm{2}{a}}{\mathrm{2}\left({x}^{\mathrm{2}\:} +\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow{ax}^{\mathrm{2}} −{a}\:=\mathrm{1}\:\Rightarrow \\ $$$${a}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:=\mathrm{1}\:\Rightarrow\:{a}\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:\left({we}\:{suppose}\:{x}\neq\overset{−} {+}\mathrm{1}\:{and}\:{x}\neq\mathrm{0}\right)\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\left\{\frac{{u}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{u}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\right\}\Rightarrow{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:{F}\left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\left\{\:\int_{\mathrm{0}} ^{{x}} \:\:\frac{{udu}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:−\int_{\mathrm{0}} ^{{x}} \:\:\frac{{udu}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\right\}\:\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\frac{{udu}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{{x}} \:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:\frac{{udu}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:=_{{u}\:={x}\alpha} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{x}\alpha\:{xd}\alpha}{{x}^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\alpha{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }\:=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\frac{{ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:\Rightarrow{f}\left({x}\right)\:=\int\:\:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{dx}\:−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\mathrm{1}}\:+{c} \\ $$$$…..{be}\:{continued}…. \\ $$$$ \\ $$