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Question Number 91271 by mathmax by abdo last updated on 29/Apr/20
calculate ∫_0 ^1 ((ln(1+x))/((x+1)^2 ))dx
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 29/Apr/20
I =∫_0 ^1 ((ln(1+x))/((x+1)^2 ))dx by psrts we get I =[−(1/(x+1))ln(x+1)]_0 ^1 +∫_0 ^1 ((ln(x+1))/(x+1))dx =−(1/2)ln(2) +∫_1 ^2 ((lnt)/t) dt (x+1=t) ∫_1 ^2 ((ln(t))/t) dt =_(lnt =u) ∫_0 ^(ln(2)) (u/e^u )e^u du =∫_0 ^(ln(2)) u du=[(u^2 /2)]_0 ^(ln(2)) =(1/2)ln^2 (2) ⇒ I =−(1/2)ln(2)+(1/2)ln^2 (2).
$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\:{by}\:{psrts}\:{we}\:{get}\: \\ $$$${I}\:=\left[−\frac{\mathrm{1}}{{x}+\mathrm{1}}{ln}\left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{lnt}}{{t}}\:{dt}\:\:\:\:\:\:\:\:\:\:\:\left({x}+\mathrm{1}={t}\right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{ln}\left({t}\right)}{{t}}\:{dt}\:=_{{lnt}\:={u}} \:\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}\right)} \:\frac{{u}}{{e}^{{u}} }{e}^{{u}} \:{du}\:\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}\right)} \:{u}\:{du}=\left[\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\:\Rightarrow\:{I}\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right). \\ $$
Answered by  M±th+et+s last updated on 29/Apr/20
I=∫_0 ^1 ((ln(1+x))/(x^2 +1))dx put x=tan(u) dx=sec^2 (u) =∫_0 ^(π/4) ((ln(tan(u)+1))/(tan^2 u+1))sec^2 (u)du =∫_0 ^(π/4) ln(tan(u)+1)du =∫_0 ^(π/4) ln(((sinu+cosu)/(cosu)))du ∫_0 ^(π/4) ln(sinu+cosu)−ln(cosu) du ∫_0 ^(π/4) ln((√2)(cos((π/4)−u))−ln(cosu) du ∫_0 ^(π/4) ln((√2))+ln(cos((π/4)−u))−ln(cosu)du J=∫_0 ^(π/4) ln(cos((π/4)−u))du (π/4)−u=t du=−dt then J=∫_(π/4) ^0 −ln(cos(t))dt∫_0 ^(π/4) ln(cos(t))dt ∫_0 ^(π/4) ln(cos(t))dt=∫_0 ^(π/4) ln(cos(u))du then ∫_0 ^(π/4) ln((√2))+ln(cos(u))−ln(cos(u))du =∫_0 ^(π/4) ln((√2))du=ln(√2)(π/4)=(π/8)ln(2)
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${put}\:{x}={tan}\left({u}\right)\:\:\:\:{dx}={sec}^{\mathrm{2}} \left({u}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({tan}\left({u}\right)+\mathrm{1}\right)}{{tan}^{\mathrm{2}} {u}+\mathrm{1}}{sec}^{\mathrm{2}} \left({u}\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tan}\left({u}\right)+\mathrm{1}\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{{sinu}+{cosu}}{{cosu}}\right){du} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sinu}+{cosu}\right)−{ln}\left({cosu}\right)\:{du} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\sqrt{\mathrm{2}}\left({cos}\left(\frac{\pi}{\mathrm{4}}−{u}\right)\right)−{ln}\left({cosu}\right)\:{du}\right. \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\sqrt{\mathrm{2}}\right)+{ln}\left({cos}\left(\frac{\pi}{\mathrm{4}}−{u}\right)\right)−{ln}\left({cosu}\right){du} \\ $$$${J}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left(\frac{\pi}{\mathrm{4}}−{u}\right)\right){du} \\ $$$$\frac{\pi}{\mathrm{4}}−{u}={t}\:\:\:\:{du}=−{dt} \\ $$$${then}\:{J}=\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} −{ln}\left({cos}\left({t}\right)\right){dt}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({t}\right)\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({t}\right)\right){dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({u}\right)\right){du} \\ $$$$ \\ $$$${then}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\sqrt{\mathrm{2}}\right)+{ln}\left({cos}\left({u}\right)\right)−{ln}\left({cos}\left({u}\right)\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\sqrt{\mathrm{2}}\right){du}={ln}\sqrt{\mathrm{2}}\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Apr/20
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$
Commented by  M±th+et+s last updated on 29/Apr/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$
Commented by mathmax by abdo last updated on 29/Apr/20
look sir the integral is ∫_0 ^1 ((ln(1+x))/((1+x)^2 ))dx not ∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx any way thank for your work...
$${look}\:{sir}\:{the}\:{integral}\:\:{is}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}\:{not}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${any}\:{way}\:{thank}\:{for}\:{your}\:{work}… \\ $$
Commented by  M±th+et+s last updated on 29/Apr/20
sorry, i didn′t notice
$${sorry},\:{i}\:{didn}'{t}\:{notice} \\ $$

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