Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 92082 by mathmax by abdo last updated on 04/May/20

let f(α) =∫_0 ^1 x(√(x^2 −x+α))dx      with α>(1/4)  1) explicit  f(α)  2) calculate g(α) =∫_0 ^1  ((xdx)/(√(x^2 −x+α)))  3) find the value of intehrals ∫_0 ^1 x(√(x^2 −x+(√2)))dx snd  ∫_0 ^1  ((xdx)/(√(x^2 −x+(√2))))

$${let}\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}{dx}\:\:\:\:\:\:{with}\:\alpha>\frac{\mathrm{1}}{\mathrm{4}} \\ $$ $$\left.\mathrm{1}\right)\:{explicit}\:\:{f}\left(\alpha\right) \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:{g}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{xdx}}{\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}} \\ $$ $$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:{intehrals}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}\sqrt{{x}^{\mathrm{2}} −{x}+\sqrt{\mathrm{2}}}{dx}\:{snd} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{xdx}}{\sqrt{{x}^{\mathrm{2}} −{x}+\sqrt{\mathrm{2}}}} \\ $$

Commented bymathmax by abdo last updated on 05/May/20

1)we have x^2 −x+α =x^2 −2(x/2)+(1/4)+α−(1/4) =(x−(1/2))^2  +((4α−1)/4)  we do the changement x−(1/2) =((√(4α−1))/2)sh(t) ⇒  f(α) =∫_(−argsh((1/(√(4α−1))))) ^(argsh((1/(√(4α−1))))) {(1/2) +((√(4α−1))/2)sh(t)}((√(4α−1))/2)ch(t)×((√(4α−1))/2)ch(t)dt  =(1/8)(4α−1) ∫_(−ln((1/(√(4α−1)))+(√(1+(1/(4α−1)))))) ^(ln((1/((√(4α−1)) ))+(√(1+(1/(4α−1)))))) (1+(√(4α−1))sh(t))ch^2 t dt  ⇒((8f(α))/(4α−1)) =∫_(x(α)) ^(y(α))  ch^2 t dt  +(√(4α−1))∫_(x(α)) ^(y(α))  sh(t)ch^2 t dt  =(1/2)∫_(x(α)) ^(y(α)) (1+ch(2t))dt  +[(1/3)ch^3 t]_(x(α)) ^(y(α))   =(1/2)(y(α)−x(α))+(1/4)[sh(2t)]_(x(α)) ^(y(α))  +(1/3)[(((e^t  +e^(−t) )/2))^3 ]_(x(α)) ^(y(α))   =((y(α)−x(α))/2) +(1/8)[ e^(2t) −e^(−2t) ]_(x(α)) ^(y(α))   +(1/(24))[ (e^t  +e^(−t) )^3 ]_(x(α)) ^(y(α))   =ln((1/(√(4α−1)))+((2(√α))/(√(4α−1))))+(1/8){(((1+2(√α))/(√(4α−1))))^2 −(((1+2(√α))/(√(4α−1))))^(−2)   −( (((1+2(√α))/(√(4α−1))))^(−2) −(((1+2(√α))/(√(4α−1))))^2 } +(1/(24)){ ( (((1+2(√α))/(√(4α−1))))−(((1+2(√α))/(√(4α−1))))^(−1) )^3   −( (((1+2(√α))/(√(4α−1))))^(−1) −(((1+2(√α))/(√(4α−1)))))^3 } rest to finish the calculus...

$$\left.\mathrm{1}\right){we}\:{have}\:{x}^{\mathrm{2}} −{x}+\alpha\:={x}^{\mathrm{2}} −\mathrm{2}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\alpha−\frac{\mathrm{1}}{\mathrm{4}}\:=\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{4}} \\ $$ $${we}\:{do}\:{the}\:{changement}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}{\mathrm{2}}{sh}\left({t}\right)\:\Rightarrow \\ $$ $${f}\left(\alpha\right)\:=\int_{−{argsh}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)} ^{{argsh}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)} \left\{\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}{\mathrm{2}}{sh}\left({t}\right)\right\}\frac{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}{\mathrm{2}}{ch}\left({t}\right)×\frac{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}{\mathrm{2}}{ch}\left({t}\right){dt} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{4}\alpha−\mathrm{1}\right)\:\int_{−{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}\alpha−\mathrm{1}}}\right)} ^{{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}\:}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}\alpha−\mathrm{1}}}\right)} \left(\mathrm{1}+\sqrt{\mathrm{4}\alpha−\mathrm{1}}{sh}\left({t}\right)\right){ch}^{\mathrm{2}} {t}\:{dt} \\ $$ $$\Rightarrow\frac{\mathrm{8}{f}\left(\alpha\right)}{\mathrm{4}\alpha−\mathrm{1}}\:=\int_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \:{ch}^{\mathrm{2}} {t}\:{dt}\:\:+\sqrt{\mathrm{4}\alpha−\mathrm{1}}\int_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \:{sh}\left({t}\right){ch}^{\mathrm{2}} {t}\:{dt} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \left(\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)\right){dt}\:\:+\left[\frac{\mathrm{1}}{\mathrm{3}}{ch}^{\mathrm{3}} {t}\right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left({y}\left(\alpha\right)−{x}\left(\alpha\right)\right)+\frac{\mathrm{1}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{t}\right)\right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \:+\frac{\mathrm{1}}{\mathrm{3}}\left[\left(\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\right)^{\mathrm{3}} \right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \\ $$ $$=\frac{{y}\left(\alpha\right)−{x}\left(\alpha\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left[\:{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} \right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \:\:+\frac{\mathrm{1}}{\mathrm{24}}\left[\:\left({e}^{{t}} \:+{e}^{−{t}} \right)^{\mathrm{3}} \right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \\ $$ $$={ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}+\frac{\mathrm{2}\sqrt{\alpha}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)+\frac{\mathrm{1}}{\mathrm{8}}\left\{\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{−\mathrm{2}} \right. \\ $$ $$−\left(\:\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{−\mathrm{2}} −\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{\mathrm{2}} \right\}\:+\frac{\mathrm{1}}{\mathrm{24}}\left\{\:\left(\:\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)−\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{−\mathrm{1}} \right)^{\mathrm{3}} \right. \\ $$ $$\left.−\left(\:\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{−\mathrm{1}} −\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)\right)^{\mathrm{3}} \right\}\:{rest}\:{to}\:{finish}\:{the}\:{calculus}... \\ $$

Commented bymathmax by abdo last updated on 05/May/20

another way let find ∫ x(√(x^2  +a)) dx with a>0  ∫ x(√(x^2  +a))dt =(1/2)∫(2x)(√(x^2  +a))dt =(1/2)∫(2x)(x^2  +a)^(1/2)  dx  =(1/2)×(1/(1+(1/2)))(x^2  +a)^((1/2)+1)  +c =(1/2)×(2/3)(x^2  +a)^(3/2)  +c  =(1/3)(x^2  +a)^(3/2)  +c ⇒  ∫ x(√(x^2 −x+α))dx =∫ x(√((x−(1/2))^2 +α−(1/4)))dx  =_(x−(1/2)=t)    ∫ (t+(1/2))(√(t^2  +α−(1/4)))dt  =∫ t(√(t^2  +α−(1/4)))dt  +∫ (√(t^2  +α−(1/4)))dt  =(1/3)(t^2  +α−(1/4))^(3/2)  + ∫ (√(t^2  +α−(1/4)))dt also we must find   ∫(√(x^2  +a))dx  ch.x =(√a)sh(t) give  ∫ (√(x^2  +a))dx =∫ (√a)ch(t)(√a)ch(t)dt =a ∫ ((1+ch(2t))/2)dt  =(a/2)t +(1/4)sh(2t) +c =(a/2) argsh((x/(√a))) +(1/2)sh(t)ch(t) +c  =(a/2)ln((x/(√a)) +(√(1+(x^2 /a)))) +(1/2)×(x/(√a))(√(1+(x^2 /a)))+c  =(a/2)ln(((x+(√(x^2  +a)))/(√a))) +(x/(2a))(√(x^2  +a)) +c  =(a/2)ln(x+(√(x^2  +a)))+(x/(2a))(√(x^2  +a))  +C ⇒  ∫ (√(t^2 +α−(1/4)))dt =(1/2)(α−(1/4))ln(x+(√(t^2  +α−(1/4))))  =(t/(2(α−(1/4))))(√(t^2  +α−(1/4))) +c    with t=x−(1/2) ....

$${another}\:{way}\:{let}\:{find}\:\int\:{x}\sqrt{{x}^{\mathrm{2}} \:+{a}}\:{dx}\:{with}\:{a}>\mathrm{0} \\ $$ $$\int\:{x}\sqrt{{x}^{\mathrm{2}} \:+{a}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{2}{x}\right)\sqrt{{x}^{\mathrm{2}} \:+{a}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} \:+{a}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{dx} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}\left({x}^{\mathrm{2}} \:+{a}\right)^{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}} \:+{c}\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} \:+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{c} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} \:+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{c}\:\Rightarrow \\ $$ $$\int\:{x}\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}{dx}\:=\int\:{x}\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dx} \\ $$ $$=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}={t}} \:\:\:\int\:\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt} \\ $$ $$=\int\:{t}\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt}\:\:+\int\:\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{3}}\left({t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\:\int\:\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt}\:{also}\:{we}\:{must}\:{find}\: \\ $$ $$\int\sqrt{{x}^{\mathrm{2}} \:+{a}}{dx}\:\:{ch}.{x}\:=\sqrt{{a}}{sh}\left({t}\right)\:{give} \\ $$ $$\int\:\sqrt{{x}^{\mathrm{2}} \:+{a}}{dx}\:=\int\:\sqrt{{a}}{ch}\left({t}\right)\sqrt{{a}}{ch}\left({t}\right){dt}\:={a}\:\int\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$ $$=\frac{{a}}{\mathrm{2}}{t}\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:+{c}\:=\frac{{a}}{\mathrm{2}}\:{argsh}\left(\frac{{x}}{\sqrt{{a}}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({t}\right){ch}\left({t}\right)\:+{c} \\ $$ $$=\frac{{a}}{\mathrm{2}}{ln}\left(\frac{{x}}{\sqrt{{a}}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{x}}{\sqrt{{a}}}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}}}+{c} \\ $$ $$=\frac{{a}}{\mathrm{2}}{ln}\left(\frac{{x}+\sqrt{{x}^{\mathrm{2}} \:+{a}}}{\sqrt{{a}}}\right)\:+\frac{{x}}{\mathrm{2}{a}}\sqrt{{x}^{\mathrm{2}} \:+{a}}\:+{c} \\ $$ $$=\frac{{a}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} \:+{a}}\right)+\frac{{x}}{\mathrm{2}{a}}\sqrt{{x}^{\mathrm{2}} \:+{a}}\:\:+{C}\:\Rightarrow \\ $$ $$\int\:\sqrt{{t}^{\mathrm{2}} +\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\alpha−\frac{\mathrm{1}}{\mathrm{4}}\right){ln}\left({x}+\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}\right) \\ $$ $$=\frac{{t}}{\mathrm{2}\left(\alpha−\frac{\mathrm{1}}{\mathrm{4}}\right)}\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}\:+{c}\:\:\:\:{with}\:{t}={x}−\frac{\mathrm{1}}{\mathrm{2}}\:.... \\ $$ $$ \\ $$

Answered by Kunal12588 last updated on 04/May/20

1) ∫_0 ^1 x(√(x^2 −x+α)) dx  =(1/2)∫_0 ^1 (2x−1)(√(x^2 −x+α)) dx+(1/2)∫_0 ^1 (√(x^2 −x+α)) dx  =(1/2)∫_0 ^1 (√((x−(1/2))^2 +((4α−1)/4))) dx; ((4α−1)/4)>0  =(1/2)[(1/2)(x−(1/2))(√(x^2 −x+α))+((4α−1)/8) ln∣(x−(1/2))+(√(x^2 −x+α))∣]_0 ^1   =(1/2)[(((√α)/4)+((4α−1)/8)ln∣(1/2)+(√α)∣)−(−((√α)/4)+((4α−1)/8)ln∣−(1/2)+(√α)∣)]  =(1/2)[((√α)/2)+((4α−1)/8)ln∣((2(√α)+1)/(2(√α)−1))∣]  =((√α)/4)+((4α−1)/(16))ln∣((4α+4(√α)+1)/(4α−1))∣  2) ∫_0 ^1  ((xdx)/(√(x^2 −x+α)))  =(1/2)∫_0 ^1  (((2x−1)dx)/(√(x^2 −x+α)))+(1/2)∫_0 ^1 (dx/(√(x^2 −x+α)))  =(1/2)∫_0 ^1 (dx/(√((x−(1/2))^2 +((√((4α−1)/4)))^2 )))  =(1/2)[ln∣(x−(1/2))+(√(x^2 −x+α))∣]_0 ^1   =(1/2)[ln∣(1/2)+(√α)∣−ln∣(√α)−(1/2)∣]  =(1/2)[ln∣(((√α)+(1/2))/((√α)−(1/2)))∣]=(1/2)ln∣((2(√α)+1)/(2(√α)−1))∣=(1/2)ln∣((4α+4(√α)+1)/(4α−1))∣

$$\left.\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\:{dx} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\:{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\:{dx} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{4}}}\:{dx};\:\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{4}}>\mathrm{0} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{8}}\:{ln}\mid\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{\sqrt{\alpha}}{\mathrm{4}}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{8}}{ln}\mid\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\alpha}\mid\right)−\left(−\frac{\sqrt{\alpha}}{\mathrm{4}}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{8}}{ln}\mid−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\alpha}\mid\right)\right] \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\sqrt{\alpha}}{\mathrm{2}}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{8}}{ln}\mid\frac{\mathrm{2}\sqrt{\alpha}+\mathrm{1}}{\mathrm{2}\sqrt{\alpha}−\mathrm{1}}\mid\right] \\ $$ $$=\frac{\sqrt{\alpha}}{\mathrm{4}}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{16}}{ln}\mid\frac{\mathrm{4}\alpha+\mathrm{4}\sqrt{\alpha}+\mathrm{1}}{\mathrm{4}\alpha−\mathrm{1}}\mid \\ $$ $$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{xdx}}{\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{2}{x}−\mathrm{1}\right){dx}}{\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{4}}}\right)^{\mathrm{2}} }} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\alpha}\mid−{ln}\mid\sqrt{\alpha}−\frac{\mathrm{1}}{\mathrm{2}}\mid\right] \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{\sqrt{\alpha}+\frac{\mathrm{1}}{\mathrm{2}}}{\sqrt{\alpha}−\frac{\mathrm{1}}{\mathrm{2}}}\mid\right]=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{2}\sqrt{\alpha}+\mathrm{1}}{\mathrm{2}\sqrt{\alpha}−\mathrm{1}}\mid=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}\alpha+\mathrm{4}\sqrt{\alpha}+\mathrm{1}}{\mathrm{4}\alpha−\mathrm{1}}\mid \\ $$

Commented byturbo msup by abdo last updated on 04/May/20

thankx

$${thankx} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com