Question Number 175051 by Mathspace last updated on 17/Aug/22
$${find}\:{the}\:{value}\:{of}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{arctanx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$
Commented by mokys last updated on 20/Aug/22
![I = ∫_0 ^( ∞) ((arctan x)/(( x^2 + 1 )^3 )) dx x = tany → dx = sec^2 y dx x= 0 → y = 0 , x→∞ ⇒ y = (𝛑/2) I = ∫_0 ^( (𝛑/2)) y cos^4 y dy u = y → du = dy , dv = cos^4 y → v = ((sin4y + 8 sin2y +12y)/(32)) I = [ ((y sin4y + 8 y sin2y + 12 y^2 )/(32))]_0 ^( (𝛑/2)) − (1/(32)) ∫_0 ^( (𝛑/2)) (sin4y +8 sin2y + 12y ) dy I = [ ((3 𝛑^2 )/(32)) ] − (1/(32)) [ − ((cos4y)/4) − 4 cos2y + 6 y^2 ]_( 0) ^( (𝛑/2)) I = ((3 𝛑^2 )/(32)) − (1/(32)) [ ( −(1/4) + 4 + ((3 𝛑^2 )/2) ) − ( − (1/4) −4 )] I = ((3 𝛑^2 )/(32)) − (1/(32)) [ ((16 + 3 𝛑^2 )/2) ] = ((6 𝛑^2 − 16 − 3 𝛑^2 )/(64)) = ((3 𝛑^2 − 16)/(64)) Aldolaimy Mohammad](https://www.tinkutara.com/question/Q175131.png)
$$\boldsymbol{{I}}\:=\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\boldsymbol{{arctan}}\:\boldsymbol{{x}}}{\left(\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\mathrm{1}\:\right)^{\mathrm{3}} }\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$\boldsymbol{{x}}\:=\:\boldsymbol{{tany}}\:\rightarrow\:\boldsymbol{{dx}}\:=\:\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{y}}\:\boldsymbol{{dx}} \\ $$$$\boldsymbol{{x}}=\:\mathrm{0}\:\rightarrow\:\boldsymbol{{y}}\:=\:\mathrm{0}\:,\:\boldsymbol{{x}}\rightarrow\infty\:\Rightarrow\:\boldsymbol{{y}}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\boldsymbol{{y}}\:\boldsymbol{{cos}}^{\mathrm{4}} \boldsymbol{{y}}\:\boldsymbol{{dy}}\: \\ $$$$ \\ $$$$\boldsymbol{{u}}\:=\:\boldsymbol{{y}}\:\rightarrow\:\boldsymbol{{du}}\:=\:\boldsymbol{{dy}}\:\:\:,\:\boldsymbol{{dv}}\:=\:\boldsymbol{{cos}}^{\mathrm{4}} \boldsymbol{{y}}\:\rightarrow\:\boldsymbol{{v}}\:=\:\frac{\boldsymbol{{sin}}\mathrm{4}\boldsymbol{{y}}\:+\:\mathrm{8}\:\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{y}}\:+\mathrm{12}\boldsymbol{{y}}}{\mathrm{32}}\: \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\left[\:\frac{\boldsymbol{{y}}\:\boldsymbol{{sin}}\mathrm{4}\boldsymbol{{y}}\:+\:\mathrm{8}\:\boldsymbol{{y}}\:\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{y}}\:+\:\mathrm{12}\:\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{32}}\underset{\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\right]}}−\:\frac{\mathrm{1}}{\mathrm{32}}\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \left(\boldsymbol{{sin}}\mathrm{4}\boldsymbol{{y}}\:+\mathrm{8}\:\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{y}}\:+\:\mathrm{12}\boldsymbol{{y}}\:\right)\:\boldsymbol{{dy}} \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\left[\:\frac{\mathrm{3}\:\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{32}}\:\right]\:−\:\frac{\mathrm{1}}{\mathrm{32}}\:\left[\:−\:\frac{\boldsymbol{{cos}}\mathrm{4}\boldsymbol{{y}}}{\mathrm{4}}\:−\:\mathrm{4}\:\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{y}}\:+\:\mathrm{6}\:\boldsymbol{{y}}^{\mathrm{2}} \:\underset{\:\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\right]}} \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\frac{\mathrm{3}\:\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{32}}\:−\:\frac{\mathrm{1}}{\mathrm{32}}\:\left[\:\left(\:−\frac{\mathrm{1}}{\mathrm{4}}\:+\:\mathrm{4}\:+\:\frac{\mathrm{3}\:\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{2}}\:\right)\:−\:\left(\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:−\mathrm{4}\:\right)\right] \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\frac{\mathrm{3}\:\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{32}}\:−\:\frac{\mathrm{1}}{\mathrm{32}}\:\left[\:\:\frac{\mathrm{16}\:+\:\mathrm{3}\:\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{2}}\:\right]\:=\:\frac{\mathrm{6}\:\boldsymbol{\pi}^{\mathrm{2}} \:−\:\mathrm{16}\:−\:\mathrm{3}\:\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{64}}\:=\:\frac{\mathrm{3}\:\boldsymbol{\pi}^{\mathrm{2}} \:−\:\mathrm{16}}{\mathrm{64}} \\ $$$$ \\ $$$${Aldolaimy}\:{Mohammad} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 20/Aug/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$