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Question Number 44179 by abdo.msup.com last updated on 22/Sep/18
1) find ∫ (dx/( (√(x^2 +x+1))+(√(x^2 −x+1)))) 2)calculate ∫_0 ^1 (dx/( (√(x^2 +x+1))+(√(x^2 −x +1))))
$$\left.\mathrm{1}\right)\:{find}\:\int\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}} \\ $$$$\left.\mathrm{2}\right){calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}} \\ $$
Commented by maxmathsup by imad last updated on 07/Oct/18
let A =∫ (dx/( (√(x^2 +x+1)) +(√(x^2 −x+1)))) ⇒A =∫ (((√(x^2 +x+1))−(√(x^2 −x+1)))/(x^2 +x+1−x^2 +x−1))dx =(1/2) ∫ ((√(x^2 +x+1))/x)dx−(1/2) ∫ ((√(x^2 −x+1))/x)dx⇒2A= H −K H = ∫ ((x^2 +x +1)/(x(√(x^2 +x+1))))dx = ∫ (x/( (√(x^2 +x+1))))dx + ∫ (dx/( (√(x^2 +x+1)))) +∫ (dx/(x(√(x^2 +x +1)))) but ∫ (x/( (√(x^2 +x +1))))dx =(1/2)∫ ((2x+1−1)/( (√(x^2 +x+1))))dx =(√(x^2 +x +1))−(1/2) ∫ (dx/( (√(x^2 +x+1)))) ⇒H =(√(x^2 +x+1))+(1/2)∫ (dx/( (√(x^2 +x+1)))) +∫ (dx/(x(√(x^2 +x+1)))) ∫ (dx/( (√(x^2 +x+1)))) =∫ (dx/( (√((x+(1/2))^2 +(3/4))))) =_(x+(1/2)=((√3)/2)tan(t)) ((√3)/2) ∫ ((1+tan^2 t)/(((√3)/2)(√(1+tan^2 t))))dt =∫ (√(1+tan^2 t))dt = ∫ (dt/(cost)) =_(tan((t/2))=u) ∫ (1/((1−u^2 )/(1+u^2 ))) ((2du)/(1+u^2 )) =∫ ((2du)/(1−u^2 )) =∫ ( (1/(1−u))+(1/(1+u)))du =ln∣((1+u)/(1−u))∣ =ln∣((1+tan((t/2)))/(1−tan((t/2))))∣ =ln∣ tan((π/4)+(t/2))∣ =ln∣tan( (1/2)arctan(((2x+1)/( (√3))))+(π/4))∣ ∫ (dx/(x(√(x^2 +x+1)))) =∫ (dx/(x(√((x+(1/2))^2 +(3/4))))) = _(x+(1/2)=((√3)/2)sh(t)) ((√3)/2)∫ ((ch(t))/((((√3)/2)sh(t)−(1/2))((√3)/2)ch(t)))dt = ∫ ((2dt)/( (√3)((e^t −e^(−t) )/2)−1)) =∫ ((4dt)/( (√3)e^t −(√3)e^(−t) −2)) =_(e^t =u) ∫ (4/( (√3)u −(√3)u^(−1) −2)) (du/u) = ∫ ((4du)/( (√3)u^2 −(√3)−2u))du =∫ ((4du)/( (√3)u^2 −2u −(√3))) Δ^′ =1+3 =4 ⇒u_1 =((1 +2)/( (√3))) =(√3) and u_2 =((1−2)/( (√3))) =−(1/( (√3))) ∫ F(u)du = ∫ ((4du)/( (√3)(u−(√3))(u+(1/( (√3)))))) =(4/( (√3))) (1/( (√3)+(1/( (√3))))) ∫ ((1/(u−(√3))) −(1/(u+(1/( (√3))))))du =ln∣((u−(√3))/(u+(1/( (√3)))))∣ =ln∣ ((e^t −(√3))/(e^t +(1/( (√3)))))∣ but t =argsh(((2x+1)/( (√3))))=ln(((2x+1)/( (√3))) +(√(1+(((2x+1)/( (√3))))^2 ))) ⇒ ∫ F(u)du =ln∣ ((((2x+1)/( (√3)))+(√(1+(((2x+1)/( (√3))))^2 ))−(√3))/(((2x+1)/( (√3)))+(√(1+(((2x+1)/( (√3))))^2 ))+(1/( (√3)))))∣ +c ⇒ H =(√(x^2 +x+1)) +(1/2)ln∣tan((1/2)arctan(((2x+1)/( (√3)))) +(π/4))∣ +ln∣ ((((2x+1)/( (√3)))+(√(1+(((2x+1)/( (√3))))^2 ))−(√3))/(((2x+1)/( (√3)))+(√(1+(((2x+1)/( (√3))))^2 ))+(1/( (√3)))))∣ +c for determination of K we use the changement x=−t ...
$${let}\:{A}\:=\int\:\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:\Rightarrow{A}\:=\int\:\:\frac{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}−{x}^{\mathrm{2}} +{x}−\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{{x}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{{x}}{dx}\Rightarrow\mathrm{2}{A}=\:{H}\:−{K} \\ $$$${H}\:=\:\int\:\:\frac{{x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}\:=\:\int\:\:\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}\:\:+\:\int\:\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:+\int\:\:\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}}} \\ $$$${but}\:\int\:\:\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{x}\:+\mathrm{1}}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{x}+\mathrm{1}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}}{dx} \\ $$$$=\sqrt{{x}^{\mathrm{2}} +{x}\:+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}}\:\Rightarrow{H}\:=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:+\int\:\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}} \\ $$$$\int\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:=\int\:\:\:\frac{{dx}}{\:\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}}\:=_{{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{tan}\left({t}\right)} \:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int\:\:\:\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {t}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}}{dt} \\ $$$$=\int\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}{dt}\:=\:\int\:\:\:\frac{{dt}}{{cost}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\:\int\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} }\:=\int\:\:\left(\:\frac{\mathrm{1}}{\mathrm{1}−{u}}+\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du}\:={ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\:={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{t}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{t}}{\mathrm{2}}\right)}\mid \\ $$$$={ln}\mid\:{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{t}}{\mathrm{2}}\right)\mid\:={ln}\mid{tan}\left(\:\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+\frac{\pi}{\mathrm{4}}\right)\mid \\ $$$$\int\:\:\:\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:\:=\int\:\:\:\frac{{dx}}{{x}\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}}\:=\:_{{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right)} \:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\int\:\:\:\:\frac{{ch}\left({t}\right)}{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right)−\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({t}\right)}{dt} \\ $$$$=\:\int\:\:\:\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{3}}\frac{{e}^{{t}} \:−{e}^{−{t}} }{\mathrm{2}}−\mathrm{1}}\:=\int\:\:\:\frac{\mathrm{4}{dt}}{\:\sqrt{\mathrm{3}}{e}^{{t}} \:−\sqrt{\mathrm{3}}{e}^{−{t}} −\mathrm{2}} \\ $$$$=_{{e}^{{t}} ={u}} \:\:\:\:\int\:\:\:\:\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}{u}\:−\sqrt{\mathrm{3}}{u}^{−\mathrm{1}} −\mathrm{2}}\:\:\frac{{du}}{{u}} \\ $$$$=\:\int\:\:\:\frac{\mathrm{4}{du}}{\:\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\sqrt{\mathrm{3}}−\mathrm{2}{u}}{du}\:=\int\:\:\:\:\frac{\mathrm{4}{du}}{\:\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\mathrm{2}{u}\:−\sqrt{\mathrm{3}}} \\ $$$$\Delta^{'} =\mathrm{1}+\mathrm{3}\:=\mathrm{4}\:\Rightarrow{u}_{\mathrm{1}} =\frac{\mathrm{1}\:+\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:=\sqrt{\mathrm{3}}\:\:{and}\:{u}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\int\:{F}\left({u}\right){du}\:=\:\int\:\:\:\:\frac{\mathrm{4}{du}}{\:\sqrt{\mathrm{3}}\left({u}−\sqrt{\mathrm{3}}\right)\left({u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)}\:=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\:\int\:\:\:\left(\frac{\mathrm{1}}{{u}−\sqrt{\mathrm{3}}}\:−\frac{\mathrm{1}}{{u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\right){du} \\ $$$$={ln}\mid\frac{{u}−\sqrt{\mathrm{3}}}{{u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\mid\:={ln}\mid\:\frac{{e}^{{t}} −\sqrt{\mathrm{3}}}{{e}^{{t}} \:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\mid\:\:{but}\:{t}\:={argsh}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)={ln}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$$\int\:{F}\left({u}\right){du}\:={ln}\mid\:\frac{\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }−\sqrt{\mathrm{3}}}{\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\mid\:+{c}\:\:\Rightarrow \\ $$$${H}\:=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+\frac{\pi}{\mathrm{4}}\right)\mid\: \\ $$$$+{ln}\mid\:\frac{\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }−\sqrt{\mathrm{3}}}{\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\mid\:+{c} \\ $$$${for}\:{determination}\:{of}\:{K}\:{we}\:{use}\:{the}\:{changement}\:{x}=−{t}\:… \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18
∫(dx/( (√(x^2 +x+1)) +(√(x^2 −x+1)) )) ∫(((√(x^2 +x+1)) −(√(x^2 −x+1)) )/((x^2 +x+1)−(x^2 −x+1)))dx ∫(((√(x^2 +x+1)) )/(2x))dx−∫(((√(x^2 −x+1)) )/(2x))dx (I_1 /2)−(I_2 /2)=I I_1 =∫((x^2 +x+1)/(x(√(x^2 +x+1))))dx =∫(x/( (√(x^2 +x+1))))dx+∫(dx/( (√(x^2 +x+1))))+∫(dx/(x(√(x^2 +x+1)))) ★(1/2)∫((2x+1−1)/( (√(x^2 +x+1))))dx+★★∫(dx/( (√(x^2 +2.x.(1/2)+(1/4)+1−(1/4)))))+★★★this part solving separTely ★★★∫(dx/(x(√(x^2 +x+1)))) x=(1/t) dx=−(1/t^2 )dt ∫((−dt)/(t^2 ×(1/t)×(√((1/t^2 )+(1/t)+1)))) ∫((−t×dt)/(t(√(t^2 +t+1)))) −∫(dt/( (√((t+(1/2))^2 +(3/4)))))=−ln∣(t+(1/2))+(√((t+(1/2))^2 +(3/4))) ∣ =−ln∣((1/x)+(1/2))+(√(((1/x)+(1/2))^2 +(3/4))) ∣ now ★ (1/2)∫((2x+1−1)/( (√(x^2 +x+1)))) (1/2)∫((d(x^2 +x+1))/( (√(x^2 +x+1))))−(1/2)∫(dx/( (√((x+(1/2))^2 +(3/4))))) (1/2)×(((√(x^2 +x+1)) )/(1/2))−(1/2)ln∣(x+(1/2))+(√(x^2 +x+1)) ∣ =(√(x^2 +x+1)) −(1/2)ln∣(x+(1/2))+(√(x^2 +x+1)) ∣ now★★∫(dx/( (√((x+(1/2))^2 +(3/4))))) =ln∣(x+(1/2))+(√(x^2 +x+1)) ∣ similarly the nxt I_2 can be solved..lenthy question... ∫(((√(x^2 −x+1)) )/x)dx ∫((x^2 −x+1)/(x(√(x^2 −x+1))))dx ★∫(x/( (√(x^2 −x+1))))dx−★★∫(dx/( (√(x^2 −x+1))))+★★★∫(dx/(x(√(x^2 −x+1)))) ★ (1/2)∫((2x−1+1)/( (√(x^2 −x+1))))dx (1/2)∫((d(x^2 −x+1))/( (√(x^2 −x+1))))dx+(1/2)∫(dx/( (√((x−(1/2))^2 +(3/4))))) (1/2)×((√(x^2 −x+1))/(1/2))+(1/2)ln∣(x−(1/2))+(√(x^2 −x+1)) ∣ ★[★ ∫(dx/( (√(x^2 −x+1)))) ∫(dx/( (√((x−(1/2))^2 +(3/4))))) =ln∣(x−(1/2))+(√(x^2 −x+1)) ∣ ★★★ ∫(dx/(x(√(x^2 −x+1)))) t=(1/x) −1∫(dt/t^2 )×(1/((1/t)(√((1/t^2 )−(1/t)+1)))) −∫(dt/( (√(1−t+t^2 )))) −∫(dt/( (√((t−(1/2))^2 +(3/4))))) −ln∣(t−(1/2))+(√(t^2 −t+1)) ∣ −ln∣((1/x)−(1/2))+(√((1/x^2 )−(1/x)+1)) now pls add ...to get ...I=((I_1 −I_2 )/2)
$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:} \\ $$$$\int\frac{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx} \\ $$$$\int\frac{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:}{\mathrm{2}{x}}{dx}−\int\frac{\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:}{\mathrm{2}{x}}{dx} \\ $$$$\frac{{I}_{\mathrm{1}} }{\mathrm{2}}−\frac{{I}_{\mathrm{2}} }{\mathrm{2}}={I} \\ $$$${I}_{\mathrm{1}} =\int\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$$$=\int\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}+\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}+\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}} \\ $$$$\bigstar\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{1}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}+\bigstar\bigstar\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}.{x}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}}+\bigstar\bigstar\bigstar{this}\:{part}\:{solving}\:{separTely} \\ $$$$\bigstar\bigstar\bigstar\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:\:{x}=\frac{\mathrm{1}}{{t}}\:\:{dx}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{−{dt}}{{t}^{\mathrm{2}} ×\frac{\mathrm{1}}{{t}}×\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{{t}}+\mathrm{1}}} \\ $$$$\int\frac{−{t}×{dt}}{{t}\sqrt{{t}^{\mathrm{2}} +{t}+\mathrm{1}}} \\ $$$$−\int\frac{{dt}}{\:\sqrt{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}}=−{ln}\mid\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\:\mid \\ $$$$=−{ln}\mid\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\:\mid \\ $$$${now}\:\bigstar\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{1}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\:\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:}{\frac{\mathrm{1}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\mathrm{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:\mid \\ $$$$=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:\mid \\ $$$${now}\bigstar\bigstar\int\frac{{dx}}{\:\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$={ln}\mid\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:\mid \\ $$$${similarly}\:\:{the}\:{nxt}\:{I}_{\mathrm{2}} \:{can}\:{be}\:{solved}..{lenthy}\:{question}… \\ $$$$\int\frac{\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:}{{x}}{dx} \\ $$$$\int\frac{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{{x}\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{dx} \\ $$$$\bigstar\int\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{dx}−\bigstar\bigstar\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}+\bigstar\bigstar\bigstar\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}} \\ $$$$\bigstar \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\:\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\mid \\ $$$$\bigstar\left[\bigstar\right. \\ $$$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}} \\ $$$$\int\frac{{dx}}{\:\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$={ln}\mid\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\mid \\ $$$$\bigstar\bigstar\bigstar \\ $$$$\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}} \\ $$$${t}=\frac{\mathrm{1}}{{x}} \\ $$$$−\mathrm{1}\int\frac{{dt}}{{t}^{\mathrm{2}} }×\frac{\mathrm{1}}{\frac{\mathrm{1}}{{t}}\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{{t}}+\mathrm{1}}} \\ $$$$−\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}+{t}^{\mathrm{2}} }} \\ $$$$−\int\frac{{dt}}{\:\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$−{ln}\mid\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{{t}^{\mathrm{2}} −{t}+\mathrm{1}}\:\mid \\ $$$$−{ln}\mid\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}}+\mathrm{1}}\: \\ $$$${now}\:{pls}\:{add}\:…{to}\:{get}\:…{I}=\frac{{I}_{\mathrm{1}} −{I}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$

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