Question Number 53228 by maxmathsup by imad last updated on 19/Jan/19
$$\left.\mathrm{1}\right)\:{find}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\left({ax}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:\:\:{with}\:\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left({a}\right) \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{xdx}}{\left({ax}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xdx}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}} \\ $$
Commented by Abdo msup. last updated on 20/Jan/19
![1)we have x^2 −x +1 =(x−(1/2))^2 +(3/4) changement x−(1/2) =((√3)/2)sh(t) give f(a) = ∫_(argsh(−(1/( (√3))))) ^(argsh((1/( (√3))))) (1/((a((1/2)+((√3)/2)sh(t))+1)((√3)/2)ch(t)))((√3)/2)ch(t)dt =∫_(ln(−(1/( (√3))) +(2/( (√3))))) ^(ln((1/( (√3))) +(2/( (√3))))) ((2dt)/(a +a(√3)sh(t)+2)) =∫_(ln((1/( (√3))))) ^(ln((√3))) ((2dt)/(a+a(√3)((e^t −e^(−t) )/2)+2)) = 4 ∫_(ln((1/( (√3))))) ^(ln((√3))) (dt/(2a +a(√3)(e^t −e^(−t) )+4)) =_(e^t =u) 4 ∫_(1/( (√3))) ^(√3) (1/(2a+a(√3)(u−u^(−1) )+4)) (du/u) =4 ∫_(1/( (√3))) ^(√3) (du/(2au +a(√3)u^2 −a(√3) +4u)) =4 ∫_(1/( (√3))) ^(√3) (du/(a(√3)u^2 +(2a+4)u−a(√3))) roots of p(u)=a(√3)u^2 +(2a+4)u−a(√3) Δ^′ =(a+2)^2 +3a^2 =a^2 +4a +4 +3a^2 =4a^2 +4a + u_1 =((−a−2+2(√(1+a+a^2 )))/(a(√3))) u_2 =((−a−2−2(√(1+a+a^2 )))/(a(√3))) ⇒ F(u) =(1/(a(√3)(u−u_1 )(u−u_2 ))) =(1/(a(√3)))(1/(u_1 −u_2 )){ (1/(u−u_1 )) −(1/(u−u_2 ))} =(1/(a(√3))) (1/((4(√(1+a+a^2 )))/(a(√3)))) {(1/(u−u_1 )) −(1/(u−u_2 ))} =(1/(4(√(1+a+a^2 )))){(1/(u−u_1 )) −(1/(u−u_2 ))}? ⇒ f(a) =(1/( (√(1+a+a^2 )))) [ln∣((u−u_1 )/(u−u_2 ))∣]_(1/( (√3))) ^(√3) =(1/( (√(1+a+a^2 )))){ln∣(((√3)−u_1 )/( (√3)−u_2 ))∣−ln∣(((1/( (√3)))−u_1 )/((1/( (√3)))−u_2 ))∣} =(1/( (√(1+a+a^2 )))){ln∣(((√3)−((−a−2+2(√(1+a+a^2 )))/(a(√3))))/( (√3)−((−a−2+2(√(1+a+a^2 )))/(a(√3)))))∣ −ln∣((1−(√3)((−a−2+2(√(1+a+a^2 )))/(a(√3))))/(1−(√3)((−a−2 −2(√(1+a+a^2 )))/(a(√3)))))∣ =(1/( (√(1+a+a^2 )))){ ln∣((4a+2+2(√(1+a+a^2 )))/(4a+2−2(√(1+a+a^2 ))))∣ −ln∣ ((2a(√3)+2−2(√(1+a+a^2 )))/(2a(√3)+2 +2(√(1+a+a^2 ))))∣} ⇒f(a) =(1/( (√(1+a+a^2 )))){ ln∣((2a+1+(√(1+a+a^2 )))/(2a+1−(√(1+a+a^2 ))))∣ −ln∣((a(√3)+1−(√(1+a+a^2 )))/(a(√3)+1 +(√(1+a+a^2 ))))∣} .](https://www.tinkutara.com/question/Q53320.png)
$$\left.\mathrm{1}\right){we}\:{have}\:{x}^{\mathrm{2}} −{x}\:+\mathrm{1}\:=\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:{changement} \\ $$$${x}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right)\:{give}\: \\ $$$${f}\left({a}\right)\:=\:\int_{{argsh}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)} ^{{argsh}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)} \:\:\:\:\frac{\mathrm{1}}{\left({a}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right)\right)+\mathrm{1}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({t}\right)}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({t}\right){dt} \\ $$$$=\int_{{ln}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)} ^{{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)} \:\:\:\frac{\mathrm{2}{dt}}{{a}\:+{a}\sqrt{\mathrm{3}}{sh}\left({t}\right)+\mathrm{2}} \\ $$$$=\int_{{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)} ^{{ln}\left(\sqrt{\mathrm{3}}\right)} \:\:\:\frac{\mathrm{2}{dt}}{{a}+{a}\sqrt{\mathrm{3}}\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}+\mathrm{2}} \\ $$$$=\:\mathrm{4}\:\int_{{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)} ^{{ln}\left(\sqrt{\mathrm{3}}\right)} \:\:\:\:\frac{{dt}}{\mathrm{2}{a}\:+{a}\sqrt{\mathrm{3}}\left({e}^{{t}} −{e}^{−{t}} \right)+\mathrm{4}} \\ $$$$=_{{e}^{{t}} ={u}} \:\:\:\:\mathrm{4}\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\sqrt{\mathrm{3}}} \:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{a}+{a}\sqrt{\mathrm{3}}\left({u}−{u}^{−\mathrm{1}} \right)+\mathrm{4}}\:\frac{{du}}{{u}} \\ $$$$=\mathrm{4}\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\sqrt{\mathrm{3}}} \:\:\:\:\:\:\:\frac{{du}}{\mathrm{2}{au}\:+{a}\sqrt{\mathrm{3}}{u}^{\mathrm{2}} \:−{a}\sqrt{\mathrm{3}}\:+\mathrm{4}{u}} \\ $$$$=\mathrm{4}\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\sqrt{\mathrm{3}}} \:\:\:\frac{{du}}{{a}\sqrt{\mathrm{3}}{u}^{\mathrm{2}} \:+\left(\mathrm{2}{a}+\mathrm{4}\right){u}−{a}\sqrt{\mathrm{3}}} \\ $$$${roots}\:{of}\:{p}\left({u}\right)={a}\sqrt{\mathrm{3}}{u}^{\mathrm{2}} +\left(\mathrm{2}{a}+\mathrm{4}\right){u}−{a}\sqrt{\mathrm{3}} \\ $$$$\Delta^{'} =\left({a}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \:={a}^{\mathrm{2}} +\mathrm{4}{a}\:+\mathrm{4}\:+\mathrm{3}{a}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}\:+ \\ $$$${u}_{\mathrm{1}} =\frac{−{a}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{{a}\sqrt{\mathrm{3}}} \\ $$$${u}_{\mathrm{2}} =\frac{−{a}−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{{a}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{\mathrm{1}}{{a}\sqrt{\mathrm{3}}\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{{a}\sqrt{\mathrm{3}}}\frac{\mathrm{1}}{{u}_{\mathrm{1}} −{u}_{\mathrm{2}} }\left\{\:\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{{a}\sqrt{\mathrm{3}}}\:\frac{\mathrm{1}}{\frac{\mathrm{4}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{{a}\sqrt{\mathrm{3}}}}\:\left\{\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\left\{\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right\}?\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\:\left[{ln}\mid\frac{{u}−{u}_{\mathrm{1}} }{{u}−{u}_{\mathrm{2}} }\mid\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\sqrt{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\left\{{ln}\mid\frac{\sqrt{\mathrm{3}}−{u}_{\mathrm{1}} }{\:\sqrt{\mathrm{3}}−{u}_{\mathrm{2}} }\mid−{ln}\mid\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−{u}_{\mathrm{1}} }{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−{u}_{\mathrm{2}} }\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\left\{{ln}\mid\frac{\sqrt{\mathrm{3}}−\frac{−{a}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{{a}\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{3}}−\frac{−{a}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{{a}\sqrt{\mathrm{3}}}}\mid\right. \\ $$$$−{ln}\mid\frac{\mathrm{1}−\sqrt{\mathrm{3}}\frac{−{a}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{{a}\sqrt{\mathrm{3}}}}{\mathrm{1}−\sqrt{\mathrm{3}}\frac{−{a}−\mathrm{2}\:−\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{{a}\sqrt{\mathrm{3}}}}\mid \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\left\{\:{ln}\mid\frac{\mathrm{4}{a}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{\mathrm{4}{a}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\mid\right. \\ $$$$\left.−{ln}\mid\:\frac{\mathrm{2}{a}\sqrt{\mathrm{3}}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{\mathrm{2}{a}\sqrt{\mathrm{3}}+\mathrm{2}\:+\mathrm{2}\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\mid\right\} \\ $$$$\Rightarrow{f}\left({a}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\left\{\:{ln}\mid\frac{\mathrm{2}{a}+\mathrm{1}+\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{\mathrm{2}{a}+\mathrm{1}−\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\mid\right. \\ $$$$\left.−{ln}\mid\frac{{a}\sqrt{\mathrm{3}}+\mathrm{1}−\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}{{a}\sqrt{\mathrm{3}}+\mathrm{1}\:+\sqrt{\mathrm{1}+{a}+{a}^{\mathrm{2}} }}\mid\right\}\:. \\ $$$$ \\ $$
Commented by Abdo msup. last updated on 20/Jan/19
$${we}\:{have}\:{f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{xdx}}{\left({ax}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{xdx}}{\left({ax}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}}\:=−{f}^{'} \left({a}\right)\:{rest}\:{calculus}\:{of} \\ $$$${f}^{'} \left({a}\right)..{be}\:{continued}… \\ $$
Commented by Abdo msup. last updated on 20/Jan/19
$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}}\:={f}\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2}\:+\mathrm{2}^{\mathrm{2}} }}\left\{{ln}\mid\frac{\mathrm{5}+\sqrt{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} }}{\mathrm{5}−\sqrt{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} }}\mid−{ln}\mid\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} }}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} }}\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\left\{{ln}\mid\frac{\mathrm{5}+\sqrt{\mathrm{7}}}{\mathrm{5}−\sqrt{\mathrm{7}}}\mid−{ln}\mid\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}+\sqrt{\mathrm{7}}}\mid\right\}\:. \\ $$$$ \\ $$