Question Number 30390 by mondodotto@gmail.com last updated on 21/Feb/18
Commented by abdo imad last updated on 21/Feb/18
$${we}\:{have}\:{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:+\mathrm{3}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}\:\:{tbe}\:{ch}\:.{x}+\mathrm{2}={cht}\:{give} \\ $$$${I}=\:\int\:\sqrt{{ch}^{\mathrm{2}} {t}\:−\mathrm{1}}\:{shtdt}\:=\:\int\:{sh}^{\mathrm{2}} {t}\:{dt}\:=\:\int\frac{\mathrm{1}−{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}{t}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int{ch}\left(\mathrm{2}{t}\right){dt}=\frac{{t}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right) \\ $$$$=\:\frac{{t}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{sh}\left({t}\right){ch}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}{argch}\left({x}+\mathrm{2}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}}\:\left({x}+\mathrm{2}\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{2}\right)\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}}\:\:+\lambda \\ $$
Commented by mondodotto@gmail.com last updated on 22/Feb/18
$$\mathrm{thAnx} \\ $$
Commented by mondodotto@gmail.com last updated on 22/Feb/18
$$\mathrm{what}\:\mathrm{does}\:\mathrm{it}\:\mathrm{mean}?\:\mathrm{ch}\:\mathrm{and}\:\mathrm{sh}???? \\ $$
Commented by prof Abdo imad last updated on 22/Feb/18
$${ch}\left({x}\right)=\:\frac{{e}^{{x}} \:+{e}^{−{x}} }{\mathrm{2}}\:\:,{sh}\left({x}\right)=\:\frac{{e}^{{x}} \:−{e}^{−{x}} }{\mathrm{2}} \\ $$$${th}\left({x}\right)=\:\frac{{shx}}{{chx}}\:\:\:\:\:{with}\:{x}\:{from}\:{R}\:{and}\:{the}\:{same} \\ $$$${definition}\:{on}\:{set}\:\mathbb{C}. \\ $$
Commented by mondodotto@gmail.com last updated on 22/Feb/18
$$\mathrm{now}\:\mathrm{i}\:\mathrm{undstnd}!\: \\ $$