Question Number 64429 by mathmax by abdo last updated on 17/Jul/19
$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{{t}+{x}+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\:\:\:\left({x}\:{real}\:{parametre}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicite}\:{form}\:{forf}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){detemine}\:{also}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({t}+{x}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){give}\:{f}^{\left({n}\right)} \left({x}\right)\:{at}\:{form}\:{of}\:{integrals} \\ $$$$\left.\mathrm{4}\right)\:{find}\:{the}\:{values}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{{t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\:\:{and}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({t}+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{5}\right)\:{find}\:{the}\:{values}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{{t}+\mathrm{1}\:+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\left({t}+\mathrm{1}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 20/Jul/19
![1) we have f(x) =∫_0 ^1 (dt/(t+x+(√(t^2 +1)))) changement t =sh(u) give f(x) =∫_0 ^(ln(1+(√2))) ((chu du)/(sh(u)+x +ch(u))) =∫_0 ^(ln(1+(√2))) (((e^u +e^(−u) )/2)/(((e^u −e^(−u) )/2)+x+((e^u +e^(−u) )/2)))du =∫_0 ^(ln(1+(√2))) ((e^u +e^(−u) )/(e^u −e^(−u) +2x +e^u +e^(−u) ))du =∫_0 ^(ln(1+(√2))) ((e^u +e^(−u) )/(2x +2e^u ))du =_(e^u =z) (1/2)∫_1 ^(1+(√2)) ((z +z^(−1) )/(x +z)) (dz/z) =(1/2) ∫_1 ^(1+(√2)) ((z+z^(−1) )/(xz +z^2 ))dz =(1/2) ∫_1 ^(1+(√2)) ((z^2 +1)/(z^2 (x+z)))dz let decompose F(z)=((z^2 +1)/(z^2 (x+z))) F(z) =(a/z) +(b/z^2 ) +(c/(x+z)) b =lim_(z→0) z^2 F(z) =(1/x) c =lim_(z→−x) (z+x)F(z) =((x^2 +1)/x^2 ) ⇒ F(z) =(a/z) +(1/(xz^2 )) +((x^2 +1)/(x^2 (z+x))) F(1) =(2/(x+1)) =a +(1/x) +((x^2 +1)/(x^2 (x+1))) ⇒2=(x+1)a+((x+1)/x) +((x^2 +1)/x^2 ) ⇒ 2=(x+1)a +((x^2 +x +x^2 +1)/x^2 ) =(x+1)a +((2x^2 +x+1)/x^2 ) ⇒ 2x^2 =x^2 (x+1)a +2x^2 +x+1 ⇒x^2 (x+1)a =−(x+1) ⇒ a =−(1/x^2 ) (if x≠−1) ⇒F(z) =−(1/(x^2 z)) +(1/(xz^2 )) +((x^2 +1)/(x^2 (z+x))) ⇒ 2f(x) =∫_1 ^(1+(√2)) {−(1/(x^2 z)) +(1/(xz^2 )) +((x^2 +1)/(x^2 (z+x)))}dz =−(1/x^2 ) ∫_1 ^(1+(√2)) (dz/z) +(1/x) ∫_1 ^(1+(√2)) (dz/z^2 ) +((x^2 +1)/x^2 ) ∫_1 ^(1+(√2)) (dz/(z+x)) =−(1/x^2 )[ln∣z∣]_1 ^(1+(√2)) −(1/x)[(1/z)]_1 ^(1+(√2)) +((x^2 +1)/x^2 )[ln∣z+x∣]_1 ^(1+(√2)) =((−ln(1+(√2)))/x^2 ) −(1/x)((1/(1+(√2))) −1) +((x^2 +1)/x^2 ){ln∣x+1+(√2)∣−ln∣x+1∣} ⇒ f(x)=(1/2){−((ln(1+(√2)))/x^2 ) +((√2)/x) +((x^2 +1)/x^2 )ln∣((x+1+(√2))/(x+1))∣ }](https://www.tinkutara.com/question/Q64644.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}+{x}+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\:\:{changement}\:{t}\:={sh}\left({u}\right)\:{give} \\ $$$${f}\left({x}\right)\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\:\:\frac{{chu}\:{du}}{{sh}\left({u}\right)+{x}\:+{ch}\left({u}\right)} \\ $$$$=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\:\frac{\frac{{e}^{{u}} \:+{e}^{−{u}} }{\mathrm{2}}}{\frac{{e}^{{u}} −{e}^{−{u}} }{\mathrm{2}}+{x}+\frac{{e}^{{u}} \:+{e}^{−{u}} }{\mathrm{2}}}{du} \\ $$$$=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\:\:\frac{{e}^{{u}} \:+{e}^{−{u}} }{{e}^{{u}} −{e}^{−{u}} \:+\mathrm{2}{x}\:+{e}^{{u}} \:+{e}^{−{u}} }{du}\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\frac{{e}^{{u}} \:+{e}^{−{u}} }{\mathrm{2}{x}\:+\mathrm{2}{e}^{{u}} }{du} \\ $$$$=_{{e}^{{u}} ={z}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:\:\:\:\frac{{z}\:+{z}^{−\mathrm{1}} }{{x}\:+{z}}\:\frac{{dz}}{{z}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:\:\:\frac{{z}+{z}^{−\mathrm{1}} }{{xz}\:+{z}^{\mathrm{2}} }{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{{z}^{\mathrm{2}} \left({x}+{z}\right)}{dz}\:\:{let}\:{decompose}\:{F}\left({z}\right)=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{{z}^{\mathrm{2}} \left({x}+{z}\right)} \\ $$$${F}\left({z}\right)\:=\frac{{a}}{{z}}\:+\frac{{b}}{{z}^{\mathrm{2}} }\:+\frac{{c}}{{x}+{z}} \\ $$$${b}\:={lim}_{{z}\rightarrow\mathrm{0}} {z}^{\mathrm{2}} \:{F}\left({z}\right)\:=\frac{\mathrm{1}}{{x}} \\ $$$${c}\:={lim}_{{z}\rightarrow−{x}} \:\left({z}+{x}\right){F}\left({z}\right)\:=\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({z}\right)\:=\frac{{a}}{{z}}\:+\frac{\mathrm{1}}{{xz}^{\mathrm{2}} }\:+\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} \left({z}+{x}\right)} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{2}}{{x}+\mathrm{1}}\:={a}\:+\frac{\mathrm{1}}{{x}}\:+\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}\:\Rightarrow\mathrm{2}=\left({x}+\mathrm{1}\right){a}+\frac{{x}+\mathrm{1}}{{x}}\:+\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{2}=\left({x}+\mathrm{1}\right){a}\:+\frac{{x}^{\mathrm{2}} \:+{x}\:+{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} }\:=\left({x}+\mathrm{1}\right){a}\:+\frac{\mathrm{2}{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:={x}^{\mathrm{2}} \left({x}+\mathrm{1}\right){a}\:+\mathrm{2}{x}^{\mathrm{2}} \:+{x}+\mathrm{1}\:\Rightarrow{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right){a}\:=−\left({x}+\mathrm{1}\right)\:\Rightarrow \\ $$$${a}\:=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\:\left({if}\:{x}\neq−\mathrm{1}\right)\:\Rightarrow{F}\left({z}\right)\:=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} {z}}\:+\frac{\mathrm{1}}{{xz}^{\mathrm{2}} }\:+\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} \left({z}+{x}\right)}\:\Rightarrow \\ $$$$\mathrm{2}{f}\left({x}\right)\:=\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \left\{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} {z}}\:+\frac{\mathrm{1}}{{xz}^{\mathrm{2}} }\:+\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} \left({z}+{x}\right)}\right\}{dz} \\ $$$$=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \frac{{dz}}{{z}}\:+\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \frac{{dz}}{{z}^{\mathrm{2}} }\:+\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} }\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:\frac{{dz}}{{z}+{x}} \\ $$$$=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left[{ln}\mid{z}\mid\right]_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} −\frac{\mathrm{1}}{{x}}\left[\frac{\mathrm{1}}{{z}}\right]_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:+\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} }\left[{ln}\mid{z}+{x}\mid\right]_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \\ $$$$=\frac{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{{x}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}}\:−\mathrm{1}\right)\:+\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} }\left\{{ln}\mid{x}+\mathrm{1}+\sqrt{\mathrm{2}}\mid−{ln}\mid{x}+\mathrm{1}\mid\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{−\frac{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{{x}^{\mathrm{2}} }\:+\frac{\sqrt{\mathrm{2}}}{{x}}\:+\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\mid\frac{{x}+\mathrm{1}+\sqrt{\mathrm{2}}}{{x}+\mathrm{1}}\mid\:\right\} \\ $$
Commented by mathmax by abdo last updated on 20/Jul/19
$$\left.\mathrm{2}\right)\:{theorem}\:{of}\:{derivation}\:{give}\: \\ $$$${f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\partial}{\partial{x}}\left(\frac{\mathrm{1}}{{t}+{x}+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\right){dt}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\left({t}+{x}+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}\right)^{\mathrm{2}} }\:=−{g}\left({x}\right)\:\Rightarrow \\ $$$${g}\left({x}\right)\:=−{f}^{'} \left({x}\right)\:\:\:{f}\:{is}\:{known}\:{rest}\:{to}\:{calculate}\:{f}^{'} \left({x}\right) \\ $$
Commented by mathmax by abdo last updated on 20/Jul/19
$$\left.\mathrm{3}\right){we}\:{have}\:{f}^{\left({n}\right)} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\partial^{{n}} }{\partial{x}^{{n}} }\left(\frac{\mathrm{1}}{{t}+{x}+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({t}+{x}+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}\right)^{{n}+\mathrm{1}} }{dt} \\ $$
Commented by mathmax by abdo last updated on 20/Jul/19
![4) ∫_0 ^1 (dt/(t+(√(t^2 +1)))) =_(t=sh(u)) ∫_0 ^(ln(1+(√2))) ((chu)/(sh(u)+ch(u)))du =∫_0 ^(ln(1+(√2))) ((e^u +e^(−u) )/(e^u −e^(−u) +e^u +e^(−u) )) du =∫_0 ^(ln(1+(√2))) ((e^u +e^(−u) )/(2e^u )) du =(1/2) ∫_0 ^(ln(1+(√2))) du +(1/2) ∫_0 ^(ln(1+(√2))) e^(−2u) du =((ln(1+(√2)))/2) −(1/4)[ e^(−2u) ]_0 ^(ln(1+(√2))) =((ln(1+(√2)))/2)−(1/4){ (1/((1+(√2))^2 )) −1} .](https://www.tinkutara.com/question/Q64647.png)
$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{{t}+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\:=_{{t}={sh}\left({u}\right)} \:\:\:\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \frac{{chu}}{{sh}\left({u}\right)+{ch}\left({u}\right)}{du} \\ $$$$=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\frac{{e}^{{u}} +{e}^{−{u}} }{{e}^{{u}} −{e}^{−{u}} \:+{e}^{{u}} \:+{e}^{−{u}} }\:{du}\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\frac{{e}^{{u}} \:+{e}^{−{u}} }{\mathrm{2}{e}^{{u}} }\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {du}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {e}^{−\mathrm{2}{u}} \:{du} \\ $$$$=\frac{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\left[\:{e}^{−\mathrm{2}{u}} \right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:−\mathrm{1}\right\}\:. \\ $$
Commented by mathmax by abdo last updated on 20/Jul/19
$$\left.\mathrm{5}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{{t}+\mathrm{1}+\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\:={f}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\sqrt{\mathrm{2}}\:+\mathrm{2}{ln}\mid\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\mid\right. \\ $$