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Question Number 197783 by pticantor last updated on 28/Sep/23

∫((x.arctg(x))/(x^2 +1))dx=?

$$\int\frac{{x}.\boldsymbol{{arctg}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}\boldsymbol{{dx}}=? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by EmGent last updated on 29/Sep/23

(1/2)arctg^2 (x)+c

$$\frac{\mathrm{1}}{\mathrm{2}}{arctg}^{\mathrm{2}} \left({x}\right)+{c} \\ $$

Commented by mr W last updated on 29/Sep/23

but (d/dx)((1/2)arctg^2 (x))=((arctg(x))/(x^2 +1))≠((x arctg(x))/(x^2 +1))

$${but}\:\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctg}^{\mathrm{2}} \left({x}\right)\right)=\frac{{arctg}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\neq\frac{{x}\:{arctg}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$

Commented by pticantor last updated on 29/Sep/23

in fact

$${in}\:{fact} \\ $$

Answered by witcher3 last updated on 30/Sep/23

x=tg(y), ⇔∫ytg(y)dy IBP =−yln(cos(y)+∫ln(cos(y))dy ∫ln(cos(y))dy=∫ln(((e^(iy) +e^(−iy) )/2))dy =z=e^(iy) ⇒dz=ie^(iy) dy,dy=(dz/(iz)) ∫ln(cos(y))dy=∫ln(((z+(1/z))/2))(dz/(iz)) =∫((ln(z^2 +1)−ln(z)−ln(2))/(iz))dz =iln(2)ln(z)+(i/2)ln^2 (z)−i∫((ln(1+z^2 ))/z)dz ∫((ln(1+z^2 ))/z)dz=∫((ln(1+z^2 ))/z^2 )zdz=(1/2)∫((ln(1+t))/t)dt =(1/2)∫((ln(1−(−t)))/((−t)))d(−t) =−(1/2)Li_2 (−t) ∫((ln(1+z^2 ))/z)=−(1/2)Li_2 (−z^2 ) ∫ln(cos(y))dy −ln(2)y−i(y^2 /2)+(i/2)Li_2 (−e^(2iy) ) ∫((xtan^(−1) (x))/(1+x^2 ))dx=xln(cos(x))−ln(2)tan^(−1) (x)−(i/2)(tan^(−1) (x))^2 +(i/2)Li_2 (−e^(2itan^(−1) (x)) )+c

$$\mathrm{x}=\mathrm{tg}\left(\mathrm{y}\right), \\ $$$$\Leftrightarrow\int\mathrm{ytg}\left(\mathrm{y}\right)\mathrm{dy}\:\mathrm{IBP} \\ $$$$=−\mathrm{yln}\left(\mathrm{cos}\left(\mathrm{y}\right)+\int\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{y}\right)\right)\mathrm{dy}\right. \\ $$$$\int\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{y}\right)\right)\mathrm{dy}=\int\mathrm{ln}\left(\frac{\mathrm{e}^{\mathrm{iy}} +\mathrm{e}^{−\mathrm{iy}} }{\mathrm{2}}\right)\mathrm{dy} \\ $$$$=\mathrm{z}=\mathrm{e}^{\mathrm{iy}} \Rightarrow\mathrm{dz}=\mathrm{ie}^{\mathrm{iy}} \mathrm{dy},\mathrm{dy}=\frac{\mathrm{dz}}{\mathrm{iz}} \\ $$$$\int\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{y}\right)\right)\mathrm{dy}=\int\mathrm{ln}\left(\frac{\mathrm{z}+\frac{\mathrm{1}}{\mathrm{z}}}{\mathrm{2}}\right)\frac{\mathrm{dz}}{\mathrm{iz}} \\ $$$$=\int\frac{\mathrm{ln}\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{z}\right)−\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{iz}}\mathrm{dz} \\ $$$$=\mathrm{iln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{z}\right)+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{z}\right)−\mathrm{i}\int\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}{\mathrm{z}}\mathrm{dz} \\ $$$$\int\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}{\mathrm{z}}\mathrm{dz}=\int\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}{\mathrm{z}^{\mathrm{2}} }\mathrm{zdz}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{ln}\left(\mathrm{1}−\left(−\mathrm{t}\right)\right)}{\left(−\mathrm{t}\right)}\mathrm{d}\left(−\mathrm{t}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{t}\right) \\ $$$$\int\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}{\mathrm{z}}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{z}^{\mathrm{2}} \right) \\ $$$$\int\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{y}\right)\right)\mathrm{dy} \\ $$$$−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{y}−\mathrm{i}\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{e}^{\mathrm{2iy}} \right) \\ $$$$\int\frac{\mathrm{xtan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\mathrm{xln}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)−\frac{\mathrm{i}}{\mathrm{2}}\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} \\ $$$$+\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{e}^{\mathrm{2itan}^{−\mathrm{1}} \left(\mathrm{x}\right)} \right)+\mathrm{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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