Question Number 68040 by mathmax by abdo last updated on 03/Sep/19
$${find}\:{f}\left({a}\right)\:=\int_{\mathrm{1}} ^{\mathrm{2}} {arctan}\left({x}+\frac{{a}}{{x}}\right){dx}\:\:{and} \\ $$$${calculate}\:{f}^{'} \left({a}\right)\:{at}\:{form}\:{of}\:{integral} \\ $$
Commented by mathmax by abdo last updated on 04/Sep/19
![f(a) =∫_1 ^2 arctan(x+(a/x))dx by parts f(a) =[x arctan(x+(a/x))]_1 ^2 −∫_1 ^2 x((1−(a/x^2 ))/(1+(x+(a/x))^2 ))dx =2 arctan(2+(a/2))−arctan(1+a)−∫_1 ^2 ((x(x^2 −a))/(x^2 +(x^2 +a)^2 ))dx let J =∫_1 ^2 ((x^3 −ax)/(x^2 +(x^2 +a)^2 ))dx ⇒ J =∫_1 ^2 ((x^3 −ax)/(x^2 +x^4 +2ax^2 +a^2 ))dx =∫_1 ^2 ((x^3 −ax)/(x^4 +(2a+1)x^2 +a^2 )) x^4 +(2a+1)x^2 +a^2 =0 ⇒t^2 +(2a+1)t +a^2 =0 with t=x^2 Δ =(2a+1)^2 −4a^2 =4a^2 +4a +1−4a^2 =4a+1 case 1 if 4a+1≥0 ⇒ t_1 =((−(2a+1)+(√(4a+1)))/2) and t_2 =((−(2a+1)−(√(4a+1)))/2) ⇒x^4 +(2a+1)x^2 +a^2 =(t−t_1 )(t−t_2 ) (x^2 −t_1 )(x^2 −t_2 ) ⇒ J =(1/(t_1 −t_2 ))∫_1 ^2 (x^3 −ax){(1/(x^2 −t_1 ))−(1/(x^2 −t_2 ))}dx =(1/(√(4a+1))){ ∫_1 ^2 ((x^3 −ax)/(x^2 −t_1 ))dx −∫_1 ^2 ((x^3 −ax)/(x^2 −t_2 ))dx} we have ∫_1 ^2 ((x^3 −ax)/(x^2 −t_1 ))dx =∫_1 ^2 ((x(x^2 −t_1 )+xt_1 −ax)/(x^2 −t_1 ))dx =[(x^2 /2)]_1 ^2 +(t_1 −a) ∫_1 ^2 ((xdx)/(x^2 −t_1 )) =2−(1/2) +((t_1 −a)/2)[ln∣x^2 −t_1 ∣]_1 ^2 =(3/2) +((t_1 −a)/2){ln∣((4−t_1 )/(1−t_1 ))∣} also we get ∫_1 ^2 ((x^3 −ax)/(x^2 −t_2 )) dx =(3/2) +((t_2 −a)/2)ln∣((4−t_2 )/(1−t_2 ))∣ ⇒ J =(1/(√(4a+1))){(((t_1 −a))/2)ln∣((4−t_1 )/(1−t_1 ))∣−((t_2 −a)/2)ln∣((4−t_2 )/(1−t_2 ))∣} ⇒ f(a) =2arctan(2+(a/2))−arctan(1+a) −(1/(√(4a+1))){ ((t_1 −a)/2)ln∣((4−t_1 )/(1−t_1 ))∣−((t_2 −a)/2)ln∣((4−t_2 )/(1−t_2 ))∣}. rest to calculate f(a) if 4a+1<0 ??....becontinued...](Q68101.png)
$${f}\left({a}\right)\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:{arctan}\left({x}+\frac{{a}}{{x}}\right){dx}\:\:{by}\:{parts}\: \\ $$$${f}\left({a}\right)\:=\left[{x}\:{arctan}\left({x}+\frac{{a}}{{x}}\right)\right]_{\mathrm{1}} ^{\mathrm{2}} −\int_{\mathrm{1}} ^{\mathrm{2}} {x}\frac{\mathrm{1}−\frac{{a}}{{x}^{\mathrm{2}} }}{\mathrm{1}+\left({x}+\frac{{a}}{{x}}\right)^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{2}\:{arctan}\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)−{arctan}\left(\mathrm{1}+{a}\right)−\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{x}\left({x}^{\mathrm{2}} −{a}\right)}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} }{dx} \\ $$$${let}\:{J}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{x}^{\mathrm{3}} −{ax}}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} }{dx}\:\Rightarrow\:{J}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{x}^{\mathrm{3}} −{ax}}{{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+\mathrm{2}{ax}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{x}^{\mathrm{3}} −{ax}}{{x}^{\mathrm{4}} \:+\left(\mathrm{2}{a}+\mathrm{1}\right){x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{4}} \:+\left(\mathrm{2}{a}+\mathrm{1}\right){x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} \:+\left(\mathrm{2}{a}+\mathrm{1}\right){t}\:+{a}^{\mathrm{2}} =\mathrm{0}\:\:{with}\:{t}={x}^{\mathrm{2}} \\ $$$$\Delta\:=\left(\mathrm{2}{a}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} \:=\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}\:+\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} =\mathrm{4}{a}+\mathrm{1} \\ $$$${case}\:\mathrm{1}\:\:{if}\:\mathrm{4}{a}+\mathrm{1}\geqslant\mathrm{0}\:\Rightarrow\:{t}_{\mathrm{1}} =\frac{−\left(\mathrm{2}{a}+\mathrm{1}\right)+\sqrt{\mathrm{4}{a}+\mathrm{1}}}{\mathrm{2}} \\ $$$${and}\:{t}_{\mathrm{2}} =\frac{−\left(\mathrm{2}{a}+\mathrm{1}\right)−\sqrt{\mathrm{4}{a}+\mathrm{1}}}{\mathrm{2}}\:\Rightarrow{x}^{\mathrm{4}} \:+\left(\mathrm{2}{a}+\mathrm{1}\right){x}^{\mathrm{2}} +{a}^{\mathrm{2}} \\ $$$$=\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)\:\left({x}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({x}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)\:\Rightarrow \\ $$$${J}\:=\frac{\mathrm{1}}{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }\int_{\mathrm{1}} ^{\mathrm{2}} \:\left({x}^{\mathrm{3}} −{ax}\right)\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{t}_{\mathrm{2}} }\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{4}{a}+\mathrm{1}}}\left\{\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{x}^{\mathrm{3}} −{ax}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }{dx}\:−\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{x}^{\mathrm{3}} −{ax}}{{x}^{\mathrm{2}} −{t}_{\mathrm{2}} }{dx}\right\}\:{we}\:{have} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{x}^{\mathrm{3}} −{ax}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }{dx}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{x}\left({x}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)+{xt}_{\mathrm{1}} −{ax}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }{dx} \\ $$$$=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{2}} \:\:+\left({t}_{\mathrm{1}} −{a}\right)\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{{xdx}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }\:=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{t}_{\mathrm{1}} −{a}}{\mathrm{2}}\left[{ln}\mid{x}^{\mathrm{2}} −{t}_{\mathrm{1}} \mid\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{{t}_{\mathrm{1}} −{a}}{\mathrm{2}}\left\{{ln}\mid\frac{\mathrm{4}−{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{1}} }\mid\right\}\:{also}\:{we}\:{get} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{x}^{\mathrm{3}} −{ax}}{{x}^{\mathrm{2}} −{t}_{\mathrm{2}} }\:{dx}\:=\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{{t}_{\mathrm{2}} −{a}}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}−{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{2}} }\mid\:\Rightarrow \\ $$$${J}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{4}{a}+\mathrm{1}}}\left\{\frac{\left({t}_{\mathrm{1}} −{a}\right)}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}−{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{1}} }\mid−\frac{{t}_{\mathrm{2}} −{a}}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}−{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{2}} }\mid\right\}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\mathrm{2}{arctan}\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)−{arctan}\left(\mathrm{1}+{a}\right) \\ $$$$−\frac{\mathrm{1}}{\sqrt{\mathrm{4}{a}+\mathrm{1}}}\left\{\:\frac{{t}_{\mathrm{1}} −{a}}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}−{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{1}} }\mid−\frac{{t}_{\mathrm{2}} −{a}}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}−{t}_{\mathrm{2}} }{\mathrm{1}−{t}_{\mathrm{2}} }\mid\right\}. \\ $$$${rest}\:{to}\:{calculate}\:{f}\left({a}\right)\:{if}\:\mathrm{4}{a}+\mathrm{1}<\mathrm{0}\:??....{becontinued}... \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 05/Sep/19
$${f}^{'} \left({a}\right)\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+\left({x}+\frac{{a}}{{x}}\right)^{\mathrm{2}} \right)}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{{x}^{\mathrm{2}} {dx}}{{x}\left\{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} \right\}} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{{xdx}}{{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+\mathrm{2}{ax}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{xdx}}{{x}^{\mathrm{4}} \:+\left(\mathrm{2}{a}+\mathrm{1}\right){x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} } \\ $$
Answered by mind is power last updated on 03/Sep/19
![integration by part f(spa)=[xarctg(x+(a/x))]−∫x(((1−(a/x^2 )))/(1+(x+(a/x))^2 ))dx =2arctg(2+(a/2))−arctg(1+a)−∫_1 ^2 ((x(x^2 −a))/(x^4 +(2a+1)x^2 +a^2 ))dx u=x^2 du=2xdx f(a)=2arctv(2+(a/2))−arctg(1+a)−∫_1 ^4 (((u−a)du)/(2(u^2 +(2a+1)u+a^2 ))) =2arctg(2+(a/2))−arctg(1+a)−∫_1 ^4 ((u+(a+(1/2))−2a−(1/2))/(2(u^2 +(2a+1)u+a^2 ))du =2arctg(2+(a/2))−arctg(1+a)−∫_1 ^4 ((u+(a+(1/2)))/(2(u^2 +(2a+1)u+a^2 ))du+((4a+1)/2)∫_1 ^4 (1/((u+((2a+1)/2))^2 +a+(1/4)))du a>−(1/4) =2arctg(2+(a/2))−arctg(1+a)−(1/4)[ln(u^2 +(2a+1)u+a^2 )]_1 ^4 +2∫_1 ^4 (du/((((2u)/(√(4a+1)))+((2a+1)/(√(4a+1))))^2 +1)) =2arctg(2+(a/2))−arctg(1+a)−(1/4)ln(((a^2 +8a+20)/(a^2 +2a+2)))+(√(4a+1))[arctg(((2u)/(√(4a+1)))+((2a+1)/(√(4a+1))))]_1 ^2 bee continued](Q68055.png)
$${integration}\:{by}\:{part} \\ $$$${f}\left({spa}\right)=\left[{xarctg}\left({x}+\frac{{a}}{{x}}\right)\right]−\int{x}\frac{\left(\mathrm{1}−\frac{{a}}{{x}^{\mathrm{2}} }\right)}{\mathrm{1}+\left({x}+\frac{{a}}{{x}}\right)^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{2}{arctg}\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)−{arctg}\left(\mathrm{1}+{a}\right)−\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}\left({x}^{\mathrm{2}} −{a}\right)}{{x}^{\mathrm{4}} +\left(\mathrm{2}{a}+\mathrm{1}\right){x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx} \\ $$$${u}={x}^{\mathrm{2}} \\ $$$${du}=\mathrm{2}{xdx} \\ $$$${f}\left({a}\right)=\mathrm{2}{arctv}\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)−{arctg}\left(\mathrm{1}+{a}\right)−\int_{\mathrm{1}} ^{\mathrm{4}} \frac{\left({u}−{a}\right){du}}{\mathrm{2}\left({u}^{\mathrm{2}} +\left(\mathrm{2}{a}+\mathrm{1}\right){u}+{a}^{\mathrm{2}} \right)} \\ $$$$=\mathrm{2}{arctg}\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)−{arctg}\left(\mathrm{1}+{a}\right)−\int_{\mathrm{1}} ^{\mathrm{4}} \frac{{u}+\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{2}{a}−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}\left({u}^{\mathrm{2}} +\left(\mathrm{2}{a}+\mathrm{1}\right){u}+{a}^{\mathrm{2}} \right.}{du} \\ $$$$=\mathrm{2}{arctg}\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)−{arctg}\left(\mathrm{1}+{a}\right)−\int_{\mathrm{1}} ^{\mathrm{4}} \frac{{u}+\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\left({u}^{\mathrm{2}} +\left(\mathrm{2}{a}+\mathrm{1}\right){u}+{a}^{\mathrm{2}} \right.}{du}+\frac{\mathrm{4}{a}+\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{4}} \frac{\mathrm{1}}{\left({u}+\frac{\mathrm{2}{a}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +{a}+\frac{\mathrm{1}}{\mathrm{4}}}{du} \\ $$$${a}>−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\mathrm{2}{arctg}\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)−{arctg}\left(\mathrm{1}+{a}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left[{ln}\left({u}^{\mathrm{2}} +\left(\mathrm{2}{a}+\mathrm{1}\right){u}+{a}^{\mathrm{2}} \right)\right]_{\mathrm{1}} ^{\mathrm{4}} +\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{4}} \frac{{du}}{\left(\frac{\mathrm{2}{u}}{\sqrt{\mathrm{4}{a}+\mathrm{1}}}+\frac{\mathrm{2}{a}+\mathrm{1}}{\sqrt{\mathrm{4}{a}+\mathrm{1}}}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\mathrm{2}{arctg}\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)−{arctg}\left(\mathrm{1}+{a}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{{a}^{\mathrm{2}} +\mathrm{8}{a}+\mathrm{20}}{{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{2}}\right)+\sqrt{\mathrm{4}{a}+\mathrm{1}}\left[{arctg}\left(\frac{\mathrm{2}{u}}{\sqrt{\mathrm{4}{a}+\mathrm{1}}}+\frac{\mathrm{2}{a}+\mathrm{1}}{\sqrt{\mathrm{4}{a}+\mathrm{1}}}\right)\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${bee}\:{continued} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 04/Sep/19
$${thanks}\:{sir}. \\ $$