Question Number 209217 by mnjuly1970 last updated on 04/Jul/24
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{calculate}}\:: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}=\:\int_{\mathrm{0}\:} ^{\:\infty} \frac{\:{tan}^{\:−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}\:+\:{x}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\: \\ $$$$ \\ $$
Answered by Berbere last updated on 04/Jul/24
![tan^(−1) (x)=u ∫_0 ^(π/2) ucos^2 (u)du=∫_0 ^(π/2) u.(((1+cos(2u))/2)) =(1/4).(π^2 /4)+(1/2)[((sin(2u))/2)u]_0 ^(π/2) −(1/4)∫sin(2u)du =(π^2 /(16))+(1/8)[cos(2u)]_0 ^(π/2) =(π^2 /(16))−(1/4)](https://www.tinkutara.com/question/Q209218.png)
$$\mathrm{tan}^{−\mathrm{1}} \left({x}\right)={u} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ucos}^{\mathrm{2}} \left({u}\right){du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {u}.\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{u}\right)}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}.\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{sin}\left(\mathrm{2}{u}\right)}{\mathrm{2}}{u}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{4}}\int{sin}\left(\mathrm{2}{u}\right){du} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{8}}\left[{cos}\left(\mathrm{2}{u}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by lepuissantcedricjunior last updated on 05/Jul/24
![i=∫_0 ^∞ ((arctan(x))/((1+x^2 )^2 ))dx posons x=tant=>dx=(1+tan^2 t)dt i=∫_0 ^(𝛑/2) (t/(1+tan^2 t))dt=∫_0 ^(π/2) tcos^2 (t)dt { ((u=t)),((v′=cos^2 (t))) :}=> { ((u′=1)),((v=(t/2)+((sin2t)/4))) :} i=[(t^2 /2)+((tsin2t)/4)]_0 ^(𝛑/2) −∫_0 ^(𝛑/2) ((t/2)+((sin2t)/4))dt =(𝛑^2 /8)−[(t^2 /4)−((cos2t)/8)]_0 ^(𝛑/2) =(𝛑^2 /8)−(𝛑^2 /(16))−(1/8)−(1/8)=(𝛑^2 /(16))−(1/4) ⇒i=∫_0 ^∞ ((arctan(x))/((1+x^2 )^2 ))dx=(𝛑^2 /(16))−(1/4)](https://www.tinkutara.com/question/Q209276.png)
$$\boldsymbol{{i}}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{2}} }\boldsymbol{{dx}} \\ $$$$\boldsymbol{{posons}}\:\boldsymbol{{x}}=\boldsymbol{{tant}}=>\boldsymbol{{dx}}=\left(\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{t}}\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{i}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \frac{\boldsymbol{{t}}}{\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{t}}}\boldsymbol{{dt}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \boldsymbol{{tcos}}^{\mathrm{2}} \left(\boldsymbol{{t}}\right)\boldsymbol{{dt}} \\ $$$$\begin{cases}{\boldsymbol{{u}}=\boldsymbol{{t}}}\\{\boldsymbol{{v}}'=\boldsymbol{{cos}}^{\mathrm{2}} \left(\boldsymbol{{t}}\right)}\end{cases}=>\begin{cases}{\boldsymbol{{u}}'=\mathrm{1}}\\{\boldsymbol{{v}}=\frac{\boldsymbol{{t}}}{\mathrm{2}}+\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{t}}}{\mathrm{4}}}\end{cases} \\ $$$$\boldsymbol{{i}}=\left[\frac{\boldsymbol{{t}}^{\mathrm{2}} }{\mathrm{2}}+\frac{\boldsymbol{{tsin}}\mathrm{2}\boldsymbol{{t}}}{\mathrm{4}}\right]_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \left(\frac{\boldsymbol{{t}}}{\mathrm{2}}+\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{t}}}{\mathrm{4}}\right)\boldsymbol{{dt}} \\ $$$$\:\:\:=\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{8}}−\left[\frac{\boldsymbol{{t}}^{\mathrm{2}} }{\mathrm{4}}−\frac{\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{t}}}{\mathrm{8}}\right]_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \\ $$$$\:\:\:=\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{8}}−\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\boldsymbol{{i}}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{2}} }\boldsymbol{{dx}}=\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$