Question Number 213003 by MrGaster last updated on 28/Oct/24
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{arctan}\:{x}\right)^{\mathrm{2}} {dx}. \\ $$$$ \\ $$
Answered by mathmax last updated on 28/Oct/24
![par parties I=[x (arctanx)^2 ]_0 ^1 −∫_0 ^1 x×((2arctanx)/(1+x^2 ))dx =(π^2 /(16))−2 ∫_0 ^1 (x/(1+x^2 )) arctan(x) dx et x=tanθ donne ∫_o ^1 ((xarctanx)/(1+x^2 ))dx=∫_0 ^(π/4) ((tanθ×θ)/(1+tan^2 θ))(1+tan^2 θ)dθ =∫_0 ^(π/4) t tant dt par partie u=t et v^′ =tant ⇒v=−ln(cost) =[−tln(cost)]_0 ^(π/4) +∫_0 ^(π/4) ln(cost)dt =−(π/4)ln((1/( (√2))))+∫_0 ^(π/4) ln(cost)dt or ∫_0 ^(π/4) ln(cost)dt=−(π/4)ln2 +(k_o /2) k_o est la constante de catalan(=Σ(((−1)^n )/((2n+1)^2 ))) ⇒ I=(π^2 /(16))−2{−(π/4)ln((1/( (√2))))−(π/4)ln2+(k_o /2)} =(π^2 /(16))−(π/4)ln2 +(π/2)ln2−k_o =(π^2 /(16))+(π/4)ln(2)−k_0](https://www.tinkutara.com/question/Q213023.png)
$${par}\:{parties}\:{I}=\left[{x}\:\left({arctanx}\right)^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {x}×\frac{\mathrm{2}{arctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{arctan}\left({x}\right)\:{dx}\:{et} \\ $$$${x}={tan}\theta\:{donne} \\ $$$$\int_{{o}} ^{\mathrm{1}} \frac{{xarctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{tan}\theta×\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {t}\:{tant}\:{dt}\:\:\:\:\:\:{par}\:{partie} \\ $$$${u}={t}\:{et}\:{v}^{'} ={tant}\:\Rightarrow{v}=−{ln}\left({cost}\right) \\ $$$$=\left[−{tln}\left({cost}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}\right){dt} \\ $$$$=−\frac{\pi}{\mathrm{4}}{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}\right){dt}\:{or} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cost}\right){dt}=−\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}\:+\frac{{k}_{{o}} }{\mathrm{2}} \\ $$$${k}_{{o}} \:{est}\:{la}\:{constante}\:{de}\:{catalan}\left(=\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\mathrm{2}\left\{−\frac{\pi}{\mathrm{4}}{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}+\frac{{k}_{{o}} }{\mathrm{2}}\right\} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}\:+\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}−{k}_{{o}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−{k}_{\mathrm{0}} \\ $$