Question Number 40146 by maxmathsup by imad last updated on 16/Jul/18
$${find}\:\:\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:−\mathrm{1}}\:+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$
Commented by maxmathsup by imad last updated on 19/Jul/18
![let I = ∫_(1/2) ^1 (dx/( (√(4x^2 −1)) −(√(4x^2 +1)))) changement 2x=t give I = ∫_1 ^2 (1/( (√(t^2 −1)) +(√(t^2 +1)))) (dt/2) ⇒ 2I = ∫_1 ^2 (((√(t^2 +1)) −(√(t^2 −1)))/2)dt ⇒4I = ∫_1 ^2 (√(t^2 +1))dt−∫_1 ^2 (√(t^2 −1))dt but ∫_1 ^2 (√(t^2 +1))dt =_(t=sh(x)) ∫_(argsh(1)) ^(argsh(2)) (√(1+sh^2 x))chxdx= ∫_(ln(1+(√2))) ^(ln(2+(√5))) ch^2 xdx =(1/2) ∫_(ln(1+(√2))) ^(ln(2+(√5))) (1+ch(2x))dx =(1/2)(ln(2+(√5))−ln(1+(√2)) +(1/4) [sh(2x)]_(ln(1+(√2))) ^(ln(2+(√5))) =(1/2){ln(2+(√5)) −ln(1+(√2))} +(1/4)[((e^(2x) −e^(−2x) )/2)]_(ln(1+(√2))) ^(ln(2+(√5))) =(1/2){ln(2+(√5))−ln(1+(√2))} +(1/8){(2+(√5))^2 −(2+(√5))^(−2) −(1+(√2))^2 +(1+(√2))^(−2) } also changement t =ch(x) give ∫_1 ^2 (√(t^2 −1))dt = ∫_(argch(1)) ^(argch(2)) sh(x)sh(x)dx = ∫_0 ^(ln(2+(√3))) ((ch(2x)−1)/2)dx=−(1/2)ln(2+(√3)) +(1/4)[sh(2x)^ ]_0 ^(ln(2+(√3))) =−(1/2)ln(2+(√3)) +(1/8){ (2+(√3))^2 −(2+(√3))^(−2) } .](https://www.tinkutara.com/question/Q40338.png)
$${let}\:{I}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}\:−\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}}\:{changement}\:\mathrm{2}{x}={t}\:{give} \\ $$$${I}\:\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\:\frac{{dt}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\:−\sqrt{{t}^{\mathrm{2}} \:−\mathrm{1}}}{\mathrm{2}}{dt}\:\Rightarrow\mathrm{4}{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}−\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:\:{but} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:\:=_{{t}={sh}\left({x}\right)} \:\:\:\int_{{argsh}\left(\mathrm{1}\right)} ^{{argsh}\left(\mathrm{2}\right)} \sqrt{\mathrm{1}+{sh}^{\mathrm{2}} {x}}{chxdx}=\:\int_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \:\:{ch}^{\mathrm{2}} {xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \left(\mathrm{1}+{ch}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:\:+\frac{\mathrm{1}}{\mathrm{4}}\:\left[{sh}\left(\mathrm{2}{x}\right)\right]_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\:−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right\}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{e}^{\mathrm{2}{x}} \:−{e}^{−\mathrm{2}{x}} }{\mathrm{2}}\right]_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right\}\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{−\mathrm{2}} \:−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{−\mathrm{2}} \right\} \\ $$$${also}\:{changement}\:\:{t}\:={ch}\left({x}\right)\:{give} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:=\:\int_{{argch}\left(\mathrm{1}\right)} ^{{argch}\left(\mathrm{2}\right)} \:{sh}\left({x}\right){sh}\left({x}\right){dx} \\ $$$$=\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \:\frac{{ch}\left(\mathrm{2}{x}\right)−\mathrm{1}}{\mathrm{2}}{dx}=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{x}\overset{} {\right)}\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{−\mathrm{2}} \right\}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18
$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:−\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}\:}{\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \mathrm{2}\sqrt{{x}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:{dx} \\ $$$${now}\:{jse}\:{formula} \\ $$$$\int\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} \:}\:\:{dx}\:\:{and}\:\int\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:{dx} \\ $$