Question Number 91842 by jagoll last updated on 03/May/20
$$\int\:\frac{{dx}}{{x}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}\:? \\ $$
Commented by mathmax by abdo last updated on 03/May/20
$${I}\:=\int\:\:\frac{{dx}}{{x}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}\:\:{let}\:{solve}\:\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{1}\:=\mathrm{0}\:\rightarrow \\ $$$$\Delta^{'} \:=\mathrm{1}+\mathrm{4}\:=\mathrm{5}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:{and}\:{x}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${I}\:=\int\:\:\frac{{dx}}{\mathrm{2}{x}\sqrt{\left({x}−{x}_{\mathrm{1}} \right)}\sqrt{{x}−{x}_{\mathrm{2}} }}\:\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{x}−{x}_{\mathrm{1}} }={t}\:\Rightarrow \\ $$$${x}−{x}_{\mathrm{1}} ={t}^{\mathrm{2}} \:\Rightarrow{I}\:=\int\:\:\frac{\mathrm{2}{tdt}}{\mathrm{2}\left({t}^{\mathrm{2}} \:+{x}_{\mathrm{1}} \right){t}\:\sqrt{{t}^{\mathrm{2}} \:+{x}_{\mathrm{1}} −{x}_{\mathrm{2}} }} \\ $$$$=\:\int\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{x}_{\mathrm{1}} \right)\sqrt{{t}^{\mathrm{2}} \:+{x}_{\mathrm{1}} −{x}_{\mathrm{2}} }}\:=\int\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\sqrt{{t}^{\mathrm{2}} +\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}} \\ $$$$=_{{t}\:=\sqrt{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}{sh}\left({t}\right)} \:\:\:\:\int\:\:\:\frac{\mathrm{1}}{\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{sh}^{\mathrm{2}} {t}\:+\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\sqrt{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}{ch}\left({t}\right)}×\sqrt{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}{ch}\left({t}\right)\:{dt} \\ $$$$=\int\:\:\:\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{5}}{sh}^{\mathrm{2}} {t}\:+\sqrt{\mathrm{5}}−\mathrm{1}}\:=\int\:\:\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{5}}×\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{5}}−\mathrm{1}} \\ $$$$=\int\:\:\frac{\mathrm{4}{dt}}{\:\sqrt{\mathrm{5}}{ch}\left(\mathrm{2}{t}\right)−\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}\:=\int\:\:\frac{\mathrm{4}{dt}}{\:\sqrt{\mathrm{5}}{ch}\left(\mathrm{2}{t}\right)+\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$$=\int\:\:\frac{\mathrm{4}{dt}}{\:\sqrt{\mathrm{5}}×\frac{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\:+\sqrt{\mathrm{5}}−\mathrm{2}}\:=\int\:\frac{\mathrm{8}{dt}}{\:\sqrt{\mathrm{5}}{e}^{\mathrm{2}{t}} \:+\sqrt{\mathrm{5}}{e}^{−\mathrm{2}{t}} \:+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}} \\ $$$$=_{{e}^{\mathrm{2}{t}} \:={z}} \:\:\:\:\mathrm{8}\:\:\int\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}{z}\:+\sqrt{\mathrm{5}}{z}^{−\mathrm{1}} \:+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}}\frac{{dz}}{\mathrm{2}{z}} \\ $$$$=\mathrm{4}\:\int\:\:\frac{{dz}}{\:\sqrt{\mathrm{5}}{z}^{\mathrm{2}} \:+\left(\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}\right){z}\:+\sqrt{\mathrm{5}}}\:\:{now}\:{its}\:{eazy}\:{to}\:{solve}… \\ $$$$ \\ $$
Answered by MJS last updated on 03/May/20
![∫(dx/(x(√(a^2 x^2 +2x−1))))= [x=(1/(1+(√(a^2 +1))sin t)) ⇔ t=arcsin ((x−1)/(x(√(a^2 +1)))) → dx=x(√(a^2 x^2 +2x−1))] =∫dt=t=arcsin ((x−1)/(x(√(a^2 +1)))) +C](https://www.tinkutara.com/question/Q91871.png)
$$\int\frac{{dx}}{{x}\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{x}=\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\mathrm{sin}\:{t}}\:\Leftrightarrow\:{t}=\mathrm{arcsin}\:\frac{{x}−\mathrm{1}}{{x}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:\rightarrow\:{dx}={x}\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}\right] \\ $$$$=\int{dt}={t}=\mathrm{arcsin}\:\frac{{x}−\mathrm{1}}{{x}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:+{C} \\ $$
Commented by jagoll last updated on 03/May/20
$${mister},\:{you}\:{always}\:{have}\: \\ $$$${an}\:{idea}\:{to}\:{solve}\:{it} \\ $$
Commented by MJS last updated on 03/May/20
I'm an integral lover ��
Commented by jagoll last updated on 03/May/20
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