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Question Number 153114 by peter frank last updated on 04/Sep/21

∫_0 ^1 ((log (1+x))/(1+x^2 ))dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{log}\:\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$

Answered by puissant last updated on 04/Sep/21

x=tan(u) → dx=1+tan^2 udu I=∫_0 ^(π/4) ((ln(1+tan(u)))/(1+tan^2 u))(1+tan^2 u)du =∫_0 ^(π/4) ln(1+tan(u))du u=(π/4)−t → du=−dt I=∫_(π/4) ^0 ln(1+tan((π/4)−t))(−dt) =∫_0 ^(π/4) ln(1+((1−tan(t))/(1+tan(t))))dt ⇒ I=∫_0 ^(π/4) ln2dt−∫_0 ^(π/4) ln(1+tan(t))dt ⇒ 2I=(π/4)ln2 ∴∵ I=(π/8)ln2..

$${x}={tan}\left({u}\right)\:\rightarrow\:{dx}=\mathrm{1}+{tan}^{\mathrm{2}} {udu} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left(\mathrm{1}+{tan}\left({u}\right)\right)}{\mathrm{1}+{tan}^{\mathrm{2}} {u}}\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left({u}\right)\right){du} \\ $$$${u}=\frac{\pi}{\mathrm{4}}−{t}\:\rightarrow\:{du}=−{dt} \\ $$$${I}=\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} {ln}\left(\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{t}\right)\right)\left(−{dt}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}−{tan}\left({t}\right)}{\mathrm{1}+{tan}\left({t}\right)}\right){dt} \\ $$$$\Rightarrow\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\mathrm{2}{dt}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left({t}\right)\right){dt} \\ $$$$\Rightarrow\:\mathrm{2}{I}=\frac{\pi}{\mathrm{4}}{ln}\mathrm{2} \\ $$$$ \\ $$$$\therefore\because\:\:{I}=\frac{\pi}{\mathrm{8}}{ln}\mathrm{2}.. \\ $$

Commented by peter frank last updated on 04/Sep/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by Ar Brandon last updated on 04/Sep/21

I=∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx, x=tanϑ =∫_0 ^(π/4) ln(1+tanϑ)dϑ =∫_0 ^(π/4) ln(cosϑ+sinϑ)dϑ−∫_0 ^(π/4) ln(cosϑ)dϑ =(1/2)(G−((πln2)/4))−((G/2)−((πln2)/4))=((πln2)/8)

$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx},\:{x}=\mathrm{tan}\vartheta \\ $$$$\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{tan}\vartheta\right)\mathrm{d}\vartheta \\ $$$$\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta+\mathrm{sin}\vartheta\right){d}\vartheta−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\vartheta\right){d}\vartheta \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({G}−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\right)−\left(\frac{{G}}{\mathrm{2}}−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\right)=\frac{\pi\mathrm{ln2}}{\mathrm{8}} \\ $$

Commented by peter frank last updated on 05/Sep/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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