Question Number 220707 by Frix last updated on 18/May/25
$$\underset{−\infty} {\overset{+\infty} {\int}}\frac{{x}\left(\mathrm{tan}^{−\mathrm{1}} \:{x}\right)^{\mathrm{3}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{e}^{\mathrm{4tan}^{−\mathrm{1}} \:{x}} \right)}{dx}=? \\ $$
Answered by SdC355 last updated on 18/May/25
![∫ ((z∙atan^3 (z))/((z^2 +1)^2 (1+e^(4∙atan(z)) ))) dz= w=atan(z) (dw/dz)=(1/(z^2 +1)) → dw=(1/(z^2 +1)) dz ∫ ((w^3 tan(w))/(sec^4 (w)(1+e^(4w) )))∙sec^2 (w) dw ∫ ((w^2 tan(w))/(sec^2 (w)(1+e^(4w) ))) dw=∫ ((w^2 sin(w)cos(w))/(1+e^(4w) )) dw= (1/2)∙∫ ((w^2 sin(2w))/(1+e^(4w) )) dw Let U be an open set in R ρ;U→R ∫_( ρ(U)) f(s) ds=∫_( U) f(ρ(w))(D∙ρ)(w) dw , D^ν f is differantial operation D^ν f=(d^ν /dt^ν )f cus w∈[−(π/2),(π/2)] (1/2) ∫_(−(π/2)) ^( +(π/2)) ((w^2 sin(2w))/(1+e^(4w) )) dw=−0.3461020688664......](https://www.tinkutara.com/question/Q220725.png)
$$\int\:\:\frac{{z}\centerdot\mathrm{atan}^{\mathrm{3}} \left({z}\right)}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+{e}^{\mathrm{4}\centerdot\mathrm{atan}\left({z}\right)} \right)}\:\mathrm{d}{z}= \\ $$$${w}=\mathrm{atan}\left({z}\right)\: \\ $$$$\frac{\mathrm{d}{w}}{\mathrm{d}{z}}=\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:\mathrm{d}{w}=\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$$$\int\:\:\frac{{w}^{\mathrm{3}} \mathrm{tan}\left({w}\right)}{\mathrm{sec}^{\mathrm{4}} \left({w}\right)\left(\mathrm{1}+{e}^{\mathrm{4}{w}} \right)}\centerdot\mathrm{sec}^{\mathrm{2}} \left({w}\right)\:\mathrm{d}{w} \\ $$$$\int\:\:\frac{{w}^{\mathrm{2}} \mathrm{tan}\left({w}\right)}{\mathrm{sec}^{\mathrm{2}} \left({w}\right)\left(\mathrm{1}+{e}^{\mathrm{4}{w}} \right)}\:\mathrm{d}{w}=\int\:\:\frac{{w}^{\mathrm{2}} \mathrm{sin}\left({w}\right)\mathrm{cos}\left({w}\right)}{\mathrm{1}+{e}^{\mathrm{4}{w}} }\:\mathrm{d}{w}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\centerdot\int\:\:\:\frac{{w}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2}{w}\right)}{\mathrm{1}+{e}^{\mathrm{4}{w}} }\:\mathrm{d}{w}\: \\ $$$$\mathrm{Let}\:{U}\:\mathrm{be}\:\mathrm{an}\:\mathrm{open}\:\mathrm{set}\:\mathrm{in}\:\mathbb{R} \\ $$$$\rho;{U}\rightarrow\mathbb{R} \\ $$$$\int_{\:\rho\left({U}\right)} \:{f}\left({s}\right)\:\mathrm{d}{s}=\int_{\:{U}} \:{f}\left(\rho\left({w}\right)\right)\left({D}\centerdot\rho\right)\left({w}\right)\:\mathrm{d}{w}\:, \\ $$$${D}^{\nu} {f}\:\mathrm{is}\:\:\mathrm{differantial}\:\mathrm{operation}\:\mathcal{D}^{\nu} {f}=\frac{\mathrm{d}^{\nu} }{\mathrm{d}{t}^{\nu} }{f} \\ $$$$\mathrm{cus}\:{w}\in\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\:+\frac{\pi}{\mathrm{2}}} \:\:\frac{{w}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2}{w}\right)}{\mathrm{1}+{e}^{\mathrm{4}{w}} }\:\mathrm{d}{w}=−\mathrm{0}.\mathrm{3461020688664}…… \\ $$
Commented by Ghisom last updated on 18/May/25
$$\mathrm{it}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:<\mathrm{0}\:\mathrm{because}\:\mathrm{the}\:\mathrm{given}\:\mathrm{function}\:>\mathrm{0} \\ $$
Answered by Ghisom last updated on 18/May/25
![I=∫_(−∞) ^(+∞) ((x(arctan x)^3 )/((x^2 +1)^2 (1+e^(4arctan x) )))dx= [x → −x] =∫_(−∞) ^(+∞) ((x(arctan x)^3 e^(4arctan x) )/((x^2 +1)^2 (1+e^(4arctan x) )))dx=J ⇒ I=((I+J)/2)=∫_(−∞) ^(+∞) ((x(arctan x)^3 )/((x^2 +1)^2 ))= [t=arctan x → dx=(x^2 +1)dt] =(1/4)∫_(−π/2) ^(π/2) t^3 sin 2t dt= [by parts] =[((3(2t^2 −1)sin 2t)/(32))−((t(2t^2 −3)cos 2t)/(16))]_(−π/2) ^(π/2) = =((π(π^2 −6))/(32))](https://www.tinkutara.com/question/Q220761.png)
$${I}=\underset{−\infty} {\overset{+\infty} {\int}}\frac{{x}\left(\mathrm{arctan}\:{x}\right)^{\mathrm{3}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{e}^{\mathrm{4arctan}\:{x}} \right)}{dx}= \\ $$$$\:\:\:\:\:\left[{x}\:\rightarrow\:−{x}\right] \\ $$$$=\underset{−\infty} {\overset{+\infty} {\int}}\frac{{x}\left(\mathrm{arctan}\:{x}\right)^{\mathrm{3}} \mathrm{e}^{\mathrm{4arctan}\:{x}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{e}^{\mathrm{4arctan}\:{x}} \right)}{dx}={J} \\ $$$$\Rightarrow \\ $$$${I}=\frac{{I}+{J}}{\mathrm{2}}=\underset{−\infty} {\overset{+\infty} {\int}}\frac{{x}\left(\mathrm{arctan}\:{x}\right)^{\mathrm{3}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arctan}\:{x}\:\rightarrow\:{dx}=\left({x}^{\mathrm{2}} +\mathrm{1}\right){dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}{t}^{\mathrm{3}} \mathrm{sin}\:\mathrm{2}{t}\:{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\left[\frac{\mathrm{3}\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{sin}\:\mathrm{2}{t}}{\mathrm{32}}−\frac{{t}\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}\right)\mathrm{cos}\:\mathrm{2}{t}}{\mathrm{16}}\right]_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} = \\ $$$$=\frac{\pi\left(\pi^{\mathrm{2}} −\mathrm{6}\right)}{\mathrm{32}} \\ $$