Question Number 150986 by talminator2856791 last updated on 17/Aug/21
$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\:\left({x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:? \\ $$$$\: \\ $$$$\: \\ $$
Answered by puissant last updated on 17/Aug/21
![Q=∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx x=tan(t)→dx=(1+tan^2 t)dt x∈[0;1]→ t∈[0;(π/4)] Q=∫_0 ^(π/4) ((ln(1+tan(t)))/((1+tan^2 t)))(1+tan^2 t)dt =∫_0 ^(π/4) ln(1+tan(t))dt t=(π/4)−u → dt=−du ⇒ Q=∫_(π/4) ^0 ln(1+tan((π/4)−u))(−du) =∫_0 ^(π/4) ln(1+tan((π/4)−u))du =∫_0 ^(π/4) ln(1+((1−tan(u))/(1+tan(u))))du =∫_0 ^(π/4) ln(((1+tan(u))/(1+tan(u)))+((1−tan(u))/(1+tan(u))))du =∫_0 ^(π/4) ln((2/(1+tan(u))))du ⇒ Q=∫_0 ^(π/4) ln2du−∫_0 ^(π/4) ln(1+tan(u))du (ln((1/A)))=−ln(A) ; ∫_a ^b f(u)du=∫_a ^b f(x)dx ⇒ 2Q=ln2∫_0 ^(π/4) du ⇒ 2Q=((πln2)/4) ∵∴ Q=((πln2)/8)... ..........Le puissant..........](https://www.tinkutara.com/question/Q150989.png)
$${Q}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${x}={tan}\left({t}\right)\rightarrow{dx}=\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$${x}\in\left[\mathrm{0};\mathrm{1}\right]\rightarrow\:{t}\in\left[\mathrm{0};\frac{\pi}{\mathrm{4}}\right] \\ $$$${Q}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left(\mathrm{1}+{tan}\left({t}\right)\right)}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)}\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left({t}\right)\right){dt} \\ $$$${t}=\frac{\pi}{\mathrm{4}}−{u}\:\rightarrow\:{dt}=−{du} \\ $$$$\Rightarrow\:{Q}=\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} {ln}\left(\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{u}\right)\right)\left(−{du}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{u}\right)\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}−{tan}\left({u}\right)}{\mathrm{1}+{tan}\left({u}\right)}\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{\mathrm{1}+{tan}\left({u}\right)}{\mathrm{1}+{tan}\left({u}\right)}+\frac{\mathrm{1}−{tan}\left({u}\right)}{\mathrm{1}+{tan}\left({u}\right)}\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{\mathrm{2}}{\mathrm{1}+{tan}\left({u}\right)}\right){du} \\ $$$$\Rightarrow\:{Q}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\mathrm{2}{du}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left({u}\right)\right){du} \\ $$$$\left({ln}\left(\frac{\mathrm{1}}{{A}}\right)\right)=−{ln}\left({A}\right)\:;\:\int_{{a}} ^{{b}} {f}\left({u}\right){du}=\int_{{a}} ^{{b}} {f}\left({x}\right){dx} \\ $$$$\Rightarrow\:\mathrm{2}{Q}={ln}\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {du} \\ $$$$\Rightarrow\:\mathrm{2}{Q}=\frac{\pi{ln}\mathrm{2}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\because\therefore\:\:{Q}=\frac{\pi{ln}\mathrm{2}}{\mathrm{8}}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:……….\mathscr{L}{e}\:{puissant}………. \\ $$
Commented by talminator2856791 last updated on 17/Aug/21
$$\:\mathrm{i}\:\mathrm{post}\:\mathrm{new}\:\mathrm{integral}. \\ $$