Question Number 42086 by maxmathsup by imad last updated on 17/Aug/18
$${let}\:\:\:\:{f}\left({x}\right)\:\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\:\:\frac{{ch}\left({t}\right)}{\mathrm{2}{xsh}\left({t}\right)\:+\mathrm{1}}\:{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\:\frac{{ch}\left({t}\right)}{\mathrm{1}+{sh}\left({t}\right)}{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\:\frac{{ch}\left({t}\right)}{\mathrm{3}{sh}\left({t}\right)\:+\mathrm{1}}{dt}\:. \\ $$
Answered by maxmathsup by imad last updated on 18/Aug/18
![1) we have f(x) = ∫_0 ^2 (((e^t +e^(−t) )/2)/(2x ((e^t −e^(−t) )/2)+1)) dt= (1/2)∫_0 ^2 ((e^t +e^(−t) )/(xe^t −x e^(−t) +1)) dt =(1/2) ∫_0 ^2 ((e^(2t) +1)/(x e^(zt) −x +et)) dt ⇒2f(x) =_(e^t =u) ∫_1 ^e^2 ((u^2 +1)/(xu^2 −x +u)) (du/u) = ∫_1 ^e^2 ((u^2 +1)/(u(xu^2 +u −x))) du let decompose F(u) =((u^2 +1)/(u(xu^2 +u−x))) poles of F Δ =1−4(−x^2 ) =1+4x^2 ⇒u_1 =((−1+(√(4x^2 +1)))/(2x)) u_2 =((−1−(√(4x^2 +1)))/(2x)) (x≠0) ⇒ F(u) =((u^2 +1)/(x u(u−u_1 )(u−u_2 ))) =(a/u) +(b/(u−u_1 )) +(c/(u−u_2 )) a =lim_(u→0) u F(u) =(1/(x u_1 u_2 )) =−(1/x) b =lim_(u→u_1 ) (u−u_1 )F(u) =((u_1 ^2 +1)/(xu_1 (u_1 −u_2 ))) c =lim_(u→u_2 ) (u−u_2 )F(u) =((u_2 ^2 +1)/(xu_(2 ) (u_2 −u_1 ))) ⇒ f(x) = ∫_1 ^e^2 F(u)du = a ∫_1 ^e^2 (du/u) +b ∫_1 ^e^2 (du/(u−u_1 )) +c ∫_1 ^e^2 (du/(u−u_2 )) =a [ln∣u∣]_1 ^e^2 +b [ln∣u−u_1 ∣]_(1 ) ^e^2 +c [ln∣u−u_2 ∣]_1 ^e^2 =2a +b (ln∣((e^2 −u_1 )/(1−u_1 ))∣) +c ln∣((e^2 −u_2 )/(1−u_2 ))∣ f(x) =((−2)/x) +(((((−1+(√(4x^2 +1)))/(2x)))^2 +1)/(x(((−1+(√(4x^2 +1)))/(2x)))(((√(4x^2 +1))/x))))ln∣ ((e^2 −((−1+(√(4x^2 +1)))/(2x)))/(1−((−1+(√(4x^2 +1)))/(2x))))∣ + (((((1+(√(4x^2 +1)))/(2x)))^2 +1)/(x(((−1−(√(4x^2 +1)))/(2x)))(−((√(4x^2 +1))/x))))ln∣((e^2 +((1+(√(4x^2 +1)))/(2x)))/(1+((1+(√(4x^2 +1)))/(2x))))∣ f(x)=((−2)/x) + (((−1+(√(4x^2 +1)))^2 +4x^2 )/(2x))ln∣((2e^2 x −(−1+(√(4x^2 +1))))/(2x−(−1+(√(4x^2 +1))))∣ −(((1+(√(4x^2 +1)))^2 +4x^2 )/(2x))ln∣ ((2e^2 x +1+(√(4x^2 +1)))/(2x+1+(√(4x^2 +1))))∣](https://www.tinkutara.com/question/Q42123.png)
$$\left.\mathrm{1}\right)\:\:{we}\:{have}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\:\:\frac{\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}}{\mathrm{2}{x}\:\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}+\mathrm{1}}\:{dt}=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\:\:\frac{{e}^{{t}} \:+{e}^{−{t}} }{{xe}^{{t}} \:−{x}\:{e}^{−{t}} \:+\mathrm{1}}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\:\frac{{e}^{\mathrm{2}{t}} \:+\mathrm{1}}{{x}\:{e}^{{zt}} \:−{x}\:+{et}}\:{dt}\:\Rightarrow\mathrm{2}{f}\left({x}\right)\:=_{{e}^{{t}} \:={u}} \:\:\:\:\int_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \:\:\:\frac{{u}^{\mathrm{2}} \:+\mathrm{1}}{{xu}^{\mathrm{2}} −{x}\:+{u}}\:\frac{{du}}{{u}} \\ $$$$=\:\int_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \:\:\:\:\:\frac{{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}\left({xu}^{\mathrm{2}} \:+{u}\:−{x}\right)}\:{du}\:\:{let}\:{decompose}\:{F}\left({u}\right)\:=\frac{{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}\left({xu}^{\mathrm{2}} +{u}−{x}\right)} \\ $$$${poles}\:{of}\:{F} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}\left(−{x}^{\mathrm{2}} \right)\:=\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \:\:\Rightarrow{u}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}} \\ $$$${u}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}}\:\:\:\:\:\:\left({x}\neq\mathrm{0}\right)\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{u}^{\mathrm{2}} \:+\mathrm{1}}{{x}\:{u}\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)}\:=\frac{{a}}{{u}}\:+\frac{{b}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{c}}{{u}−{u}_{\mathrm{2}} } \\ $$$${a}\:={lim}_{{u}\rightarrow\mathrm{0}} {u}\:{F}\left({u}\right)\:=\frac{\mathrm{1}}{{x}\:{u}_{\mathrm{1}} {u}_{\mathrm{2}} }\:\:=−\frac{\mathrm{1}}{{x}} \\ $$$${b}\:={lim}_{{u}\rightarrow{u}_{\mathrm{1}} } \left({u}−{u}_{\mathrm{1}} \right){F}\left({u}\right)\:=\frac{{u}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}}{{xu}_{\mathrm{1}} \left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right)} \\ $$$${c}\:={lim}_{{u}\rightarrow{u}_{\mathrm{2}} } \left({u}−{u}_{\mathrm{2}} \right){F}\left({u}\right)\:=\frac{{u}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}}{{xu}_{\mathrm{2}\:} \left({u}_{\mathrm{2}} −{u}_{\mathrm{1}} \right)}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\int_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \:{F}\left({u}\right){du}\:\:=\:{a}\:\int_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \:\:\frac{{du}}{{u}}\:\:+{b}\:\int_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \:\:\frac{{du}}{{u}−{u}_{\mathrm{1}} }\:\:+{c}\:\int_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \:\:\frac{{du}}{{u}−{u}_{\mathrm{2}} } \\ $$$$={a}\:\left[{ln}\mid{u}\mid\right]_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \:\:\:+{b}\:\:\left[{ln}\mid{u}−{u}_{\mathrm{1}} \mid\right]_{\mathrm{1}\:} ^{{e}^{\mathrm{2}} } \:\:\:\:\:+{c}\:\:\left[{ln}\mid{u}−{u}_{\mathrm{2}} \mid\right]_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \\ $$$$=\mathrm{2}{a}\:\:+{b}\:\:\left({ln}\mid\frac{{e}^{\mathrm{2}} \:−{u}_{\mathrm{1}} }{\mathrm{1}−{u}_{\mathrm{1}} }\mid\right)\:+{c}\:{ln}\mid\frac{{e}^{\mathrm{2}} −{u}_{\mathrm{2}} }{\mathrm{1}−{u}_{\mathrm{2}} }\mid \\ $$$${f}\left({x}\right)\:=\frac{−\mathrm{2}}{{x}}\:\:+\frac{\left(\frac{−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}}\right)^{\mathrm{2}} \:+\mathrm{1}}{{x}\left(\frac{−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}}\right)\left(\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{{x}}\right)}{ln}\mid\:\:\frac{{e}^{\mathrm{2}} −\frac{−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}}}{\mathrm{1}−\frac{−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}}}\mid \\ $$$$+\:\frac{\left(\frac{\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}}\right)^{\mathrm{2}} \:+\mathrm{1}}{{x}\left(\frac{−\mathrm{1}−\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}}\right)\left(−\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{{x}}\right)}{ln}\mid\frac{{e}^{\mathrm{2}} \:\:+\frac{\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}}}{\mathrm{1}+\frac{\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}}}\mid \\ $$$${f}\left({x}\right)=\frac{−\mathrm{2}}{{x}}\:\:+\:\:\frac{\left(−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}{x}}{ln}\mid\frac{\mathrm{2}{e}^{\mathrm{2}} {x}\:−\left(−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}\right)}{\mathrm{2}{x}−\left(−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}\right.}\mid \\ $$$$−\frac{\left(\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}{x}}{ln}\mid\:\:\frac{\mathrm{2}{e}^{\mathrm{2}} {x}\:+\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}\mid \\ $$
Commented by maxmathsup by imad last updated on 18/Aug/18
$$\mathrm{2}{f}\left({x}\right)\:=\:−\frac{\mathrm{2}}{{x}}\:+….\Rightarrow \\ $$$${f}\left({x}\right)\:=−\frac{\mathrm{1}}{{x}}\:+\frac{\left(−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}}{ln}\mid\frac{\mathrm{2}{e}^{\mathrm{2}} {x}−\left(−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}\right.}{\mathrm{2}{x}−\left(−\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\right.}\mid \\ $$$$−\frac{\left(\mathrm{1}+\sqrt{\left.\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}}{ln}\mid\:\:\frac{\mathrm{2}{e}^{\mathrm{2}} {x}\:+\mathrm{1}\:+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}{x}+\mathrm{1}\:+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}}\mid. \\ $$
Commented by math khazana by abdo last updated on 18/Aug/18
$$\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\:\frac{{ch}\left({t}\right)}{\mathrm{1}+{sh}\left({t}\right)}{dt}\:={f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\mathrm{2}\:+\frac{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{2}}{ln}\mid\:\frac{{e}^{\mathrm{2}} \:−\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{1}−\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)}\mid \\ $$$$−\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{{e}^{\mathrm{2}} \:+\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\mid \\ $$$$=−\mathrm{2}\:\:+\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){ln}\mid\frac{{e}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\mid−\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){ln}\mid\frac{{e}^{\mathrm{2}} \:+\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\mid\:. \\ $$