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Question Number 145636 by mathmax by abdo last updated on 06/Jul/21

calculate ∫_0 ^∞ ((arctanx)/((1+x^2 )^2 ))dx

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

Answered by Dwaipayan Shikari last updated on 06/Jul/21

∫_0 ^∞ ((tan^(−1) (x))/((1+x^2 )^2 ))dx x=tanθ =∫_0 ^(π/2) (θ/(sec^4 θ))sec^2 θdθ =∫_0 ^(π/2) θcos^2 θ dθ=(1/2)∫_0 ^(π/2) θ+θcos2θdθ =(π^2 /(16))−(1/4)∫_0 ^(π/2) sin2θdθ=(π^2 /(16))−(1/4)

$$\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\:\:\:\:\:\:{x}={tan}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\theta}{{sec}^{\mathrm{4}} \theta}{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \theta{cos}^{\mathrm{2}} \theta\:{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \theta+\theta{cos}\mathrm{2}\theta{d}\theta \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}\mathrm{2}\theta{d}\theta=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Answered by mathmax by abdo last updated on 07/Jul/21

Ψ=∫_0 ^∞ ((arctanx)/((1+x^2 )^2 ))dx cha7gement x=tant give Ψ=∫_0 ^(π/2) (t/((1+tan^2 t)^2 ))(1+tan^2 t)dt =∫_0 ^(π/2) (t/(1+tan^2 t))dt =∫_0 ^(π/2) tcos^2 t dt =∫_0 ^(π/2) t(((1+cos(2t))/2))dt=(1/2)∫_0 ^(π/2) tdt +(1/2)∫_0 ^(π/2) tcos(2t)dt ∫_0 ^(π/2) tdt =[(t^2 /2)]_0 ^(π/2) =(π^2 /8) ∫_0 ^(π/2) tcos(2t)dt =[(t/2)sin(2t)]_0 ^(π/2) −∫_0 ^(π/2) ((sin(2t))/2)dt =[(1/4)cos(2t)]_0 ^(π/2) =(1/4)(−2)=−(1/2) ⇒ Ψ=(π^2 /(16))−(1/4)

$$\Psi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\:\mathrm{cha7gement}\:\mathrm{x}=\mathrm{tant}\:\mathrm{give} \\ $$$$\Psi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{t}}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}\right)^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}\right)\mathrm{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{t}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}}\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tcos}^{\mathrm{2}} \mathrm{t}\:\mathrm{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{t}\left(\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2t}\right)}{\mathrm{2}}\right)\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{tdt}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tcos}\left(\mathrm{2t}\right)\mathrm{dt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{tdt}\:=\left[\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{tcos}\left(\mathrm{2t}\right)\mathrm{dt}\:=\left[\frac{\mathrm{t}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\left(\mathrm{2t}\right)}{\mathrm{2}}\mathrm{dt} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\left(\mathrm{2t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{2}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\Psi=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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