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Question Number 162792 by HongKing last updated on 01/Jan/22

Find:  𝛀  = ∫_( 0) ^( ∞)  ((arctan(x))/(x∙(x^2  - x + 1))) dx

$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:\:=\:\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{x}\centerdot\left(\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{x}\:+\:\mathrm{1}\right)}\:\mathrm{dx} \\ $$

Answered by phanphuoc last updated on 01/Jan/22

we have arctan(x)=arctan(tx)∣_0 ^1 =∫_0 ^1 ((xdt)/(1+(tx)^2 ))  Ω=∫_0 ^∞ ∫_0 ^1 ((dtdx)/((1+(tx)^2 (1−x+x^2 )))  you can find f(t),g(t)  1/(((1+(tx)^2 )(1−x+x^2 )))=((f(t))/(1+(tx)^2 ))+((g(t))/(1−x+x^2 ))

$${we}\:{have}\:{arctan}\left({x}\right)={arctan}\left({tx}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xdt}}{\mathrm{1}+\left({tx}\right)^{\mathrm{2}} } \\ $$$$\Omega=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dtdx}}{\left(\mathrm{1}+\left({tx}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)\right.} \\ $$$${you}\:{can}\:{find}\:{f}\left({t}\right),{g}\left({t}\right) \\ $$$$\mathrm{1}/\left(\left(\left(\mathrm{1}+\left({tx}\right)^{\mathrm{2}} \right)\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)\right)\right)=\frac{{f}\left({t}\right)}{\mathrm{1}+\left({tx}\right)^{\mathrm{2}} }+\frac{{g}\left({t}\right)}{\mathrm{1}−{x}+{x}^{\mathrm{2}} } \\ $$$$ \\ $$

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