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Question Number 62208 by maxmathsup by imad last updated on 17/Jun/19
find f(a) =∫  (x−a)(√(x^2  +a^2 ))dx
$${find}\:{f}\left({a}\right)\:=\int\:\:\left({x}−{a}\right)\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 18/Jun/19
f(a) =∫ (x−a)(√(x^2  +a^2 ))dx   ⇒f(a)=∫ x(√(x^2  +a^2 ))dx−a ∫(√(x^2  +a^2 ))dx  ∫ x(√(x^2  +a^2 ))dx = (1/3)(x^2  +a^2 )^(3/2)  +c_1   ∫ (√(x^2  +a^2 ))dx =_(x =ash(t))   ∫ ∣a∣ch(t)ach(t)dt =a∣a∣∫  ch^2 t dt  =a∣a∣ ∫  ((1+ch(2t))/2)dt =((a∣a∣)/2)t  +((a∣a∣)/4) sh(2t) +c_2   =((a∣a∣)/2)t   +((a∣a∣)/2)sh(t)ch(t) +c_2  =((a∣a∣)/2){ argsh((x/a)) +(x/a)(√(1+(x^2 /a^2 )))} +c_2   =((a∣a∣)/2){ln((x/a)+(√(1+(x^2 /a^2 )))) +(x/(a∣a∣))(√(a^2  +x^2 ))} +c_2   =((a∣a∣)/2) ln((x/a) +((√(a^2 +x^2 ))/(∣a∣))) +(x/2) (√(a^2  +x^2 )) +c_2  ⇒  f(a) =(1/3)(x^2  +a^2 )^(3/2)  −((a^2 ∣a∣)/2)ln((x/a) +((√(a^2  +x^2 ))/(∣a∣)))−((ax)/2)(√(a^2  +x^2 )) +C
$${f}\left({a}\right)\:=\int\:\left({x}−{a}\right)\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\:\:\:\Rightarrow{f}\left({a}\right)=\int\:{x}\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}−{a}\:\int\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx} \\ $$$$\int\:{x}\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{c}_{\mathrm{1}} \\ $$$$\int\:\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\:=_{{x}\:={ash}\left({t}\right)} \:\:\int\:\mid{a}\mid{ch}\left({t}\right){ach}\left({t}\right){dt}\:={a}\mid{a}\mid\int\:\:{ch}^{\mathrm{2}} {t}\:{dt} \\ $$$$={a}\mid{a}\mid\:\int\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:=\frac{{a}\mid{a}\mid}{\mathrm{2}}{t}\:\:+\frac{{a}\mid{a}\mid}{\mathrm{4}}\:{sh}\left(\mathrm{2}{t}\right)\:+{c}_{\mathrm{2}} \\ $$$$=\frac{{a}\mid{a}\mid}{\mathrm{2}}{t}\:\:\:+\frac{{a}\mid{a}\mid}{\mathrm{2}}{sh}\left({t}\right){ch}\left({t}\right)\:+{c}_{\mathrm{2}} \:=\frac{{a}\mid{a}\mid}{\mathrm{2}}\left\{\:{argsh}\left(\frac{{x}}{{a}}\right)\:+\frac{{x}}{{a}}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right\}\:+{c}_{\mathrm{2}} \\ $$$$=\frac{{a}\mid{a}\mid}{\mathrm{2}}\left\{{ln}\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)\:+\frac{{x}}{{a}\mid{a}\mid}\sqrt{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\right\}\:+{c}_{\mathrm{2}} \\ $$$$=\frac{{a}\mid{a}\mid}{\mathrm{2}}\:{ln}\left(\frac{{x}}{{a}}\:+\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{\mid{a}\mid}\right)\:+\frac{{x}}{\mathrm{2}}\:\sqrt{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:−\frac{{a}^{\mathrm{2}} \mid{a}\mid}{\mathrm{2}}{ln}\left(\frac{{x}}{{a}}\:+\frac{\sqrt{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }}{\mid{a}\mid}\right)−\frac{{ax}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:+{C}\: \\ $$
Commented by maxmathsup by imad last updated on 18/Jun/19
if a=0 we get  f(a) =∫ x∣x∣dx =....
$${if}\:{a}=\mathrm{0}\:{we}\:{get}\:\:{f}\left({a}\right)\:=\int\:{x}\mid{x}\mid{dx}\:=…. \\ $$
Answered by tanmay last updated on 17/Jun/19
∫x(√(x^2 +a^2 )) dx−a∫(√(x^2 +a^2  )) dx  (1/2)∫(x^2 +a^2 )^(1/2) ×d(x^2 +a^2 )−a[((x(√(x^2 +a^2 )) )/2)+(a^2 /2)ln(x+(√(x^2 +a^2 )) )]+c  (1/2)×(((x^2 +a^2 )^(3/2) )/(3/2))−a[((x(√(x^2 +a^2 )))/2)+(a^2 /2)ln(x+(√(x^2 +a^2 )) )+c
$$\int{x}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:{dx}−{a}\int\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} \:}\:{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} ×{d}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)−{a}\left[\frac{{x}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\right)\right]+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{\mathrm{3}}{\mathrm{2}}}−{a}\left[\frac{{x}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\right)+{c}\right. \\ $$

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