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Question Number 225382 by Osefavour last updated on 23/Oct/25
Been a while guys ∫_0 ^( 1) ((xln(1+x))/(1+x^2 ))dx
$$\mathrm{Been}\:\mathrm{a}\:\mathrm{while}\:\mathrm{guys} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{xln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$
Answered by peace2 last updated on 24/Oct/25
A=∫_0 ^1 ((xln(1+x))/(1+x^2 ))dx;B=∫_0 ^1 ((xln(1−x))/(1+x^2 ));A+B=∫_0 ^1 ((xln(1−x^2 ))/(1+x^2 ))dx =(1/2)∫_0 ^1 ((ln(1−t))/(1+t))dt let 1+t=2z⇒A+B=∫_(1/2) ^1 ((ln(2−2z))/(2z)).2dz =∫_(1/2) ^1 ((ln(2))/z)dz+((ln(1−z))/z)dz=ln^2 (2)−Li_2 (1)+Li_2 ((1/2)) =ln^2 (2)−(𝛑^2 /6)+Li_2 ((1/2))p B−A=∫_0 ^1 ((xln(((1−x)/(1+x))))/(1+x^2 ))dx;t=((1−x)/(1+x))⇒B−A=∫_0 ^1 ((((1−t)/(1+t)).ln(t))/((2+2t^2 )/((1+t)^2 ))).(2/((1+t)^2 ))dt =∫_0 ^1 (((1−t)ln(t))/((1+t)(1+t^2 )))dt=∫_0 ^1 ((1+t^2 −t(1+t))/((1+t^2 )(1+t)))ln(t)dt =∫_0 ^1 ((ln(t))/(1+t))dt−∫_0 ^1 ((tln(t))/(1+t^2 ))dt;t^2 →t =(3/4)∫_0 ^1 ((ln(t))/(1+t))=(3/4)Σ_(n≥0) ∫_0 ^1 (−1)^n t^n ln(t)dt =(3/4)−Σ_(n≥0) (((−1)^n )/((n+1)^2 ))=(3/4)Σ_(n≥1) (((−1)^n )/n^2 )=(3/4){.−(1/2)ζ(2)} =−((3π^2 )/(48)) A=(1/2)(A+B−(B−A))=(1/2)(ln^2 (2)−(π^2 /6)+Li_2 ((1/2))+((3π^2 )/(48))) ((ln^2 (2))/2)−((5π^2 )/(96))+((Li_2 ((1/2)))/2)
$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx};{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} };{A}+{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}+{t}}{dt}\:{let}\:\mathrm{1}+{t}=\mathrm{2}{z}\Rightarrow{A}+{B}=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{2}−\mathrm{2}{z}\right)}{\mathrm{2}{z}}.\mathrm{2}{dz} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{2}\right)}{{z}}{dz}+\frac{{ln}\left(\mathrm{1}−{z}\right)}{{z}}{dz}={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{L}{i}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}}+{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right){p} \\ $$$${B}−{A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx};{t}=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\Rightarrow{B}−{A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}.{ln}\left({t}\right)}{\frac{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }}.\frac{\mathrm{2}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{t}\right){ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+{t}^{\mathrm{2}} −{t}\left(\mathrm{1}+{t}\right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}\right)}{ln}\left({t}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\mathrm{1}+{t}}{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt};{t}^{\mathrm{2}} \rightarrow{t} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\mathrm{1}+{t}}=\frac{\mathrm{3}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} {t}^{{n}} {ln}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}−\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{4}}\left\{.−\frac{\mathrm{1}}{\mathrm{2}}\zeta\left(\mathrm{2}\right)\right\} \\ $$$$=−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left({A}+{B}−\left({B}−{A}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{48}}\right) \\ $$$$\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{96}}+\frac{{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

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