Question Number 206579 by York12 last updated on 19/Apr/24
$$\mathrm{Evaluate}\::\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}}{d}\left({x}\right). \\ $$
Answered by Berbere last updated on 19/Apr/24
![=[ln(1+x)ln(1+x^2 )]_0 ^1 −∫_0 ^1 ((ln(1+x)2x)/(1+x^2 ))dx =ln^2 (2)−∫_0 ^1 ((ln(1+x)2x)/(1+x^2 ))=ln^2 (2)−A B=∫_0 ^1 ((ln(1−x))/(1+x^2 ))2x,A+B=∫_0 ^1 ((2xln(1−x^2 ))/(1+x^2 ))dx =∫_0 ^1 ((ln(1−x))/(1+x))dx=∫_0 ^1 ((ln(2)+ln((1/2)−(x/2)))/(2((1/2)+(x/2))))dx =ln(2)ln(1+x)]_0 ^1 +∫_0 ^1 ((ln(1−((1/2)+(x/2)))/((1/2)+(x/2)))d((x/2)+(1/2))=ln^2 (2)−Li_2 (((x+1)/2))]_0 ^1 =ln^2 (2)−Li_2 (1)+Li_2 ((1/2)) B−A=∫_0 ^1 ((2xln(((1−x)/(1+x))))/(1+x^2 ))dx;t=((1−x)/(1+x)) ∫_0 ^1 ((2(1−t))/(1+t)).((ln(t))/((2(1+t^2 ))/((1+t)^2 ))).((2dt)/((1+t)^2 ))=2∫_0 ^1 (((1−t)(ln(t))/((1+t)(1+t^2 ))) =2∫_0 ^1 ((ln(t))/(1+t))+((−t)/(t^2 +1))ln(t)=−2∫_0 ^1 ((ln(1+t))/t)+∫_0 ^1 ((ln(1+t^2 ))/t)dt_(t^2 →t) =−2∫_0 ^1 ((ln(1+t))/t)+(1/2)∫_0 ^1 ((ln(1+t))/t)dt=−(3/2)∫_0 ^1 ((ln(1−(−t)))/((−t)))d(−t)= (3/2)Li_2 (−t)]_0 ^1 =((3Li_2 (−1))/2) A=(1/2)((B+A)−(B−A)) =(1/2)(ln^2 (2)−Li_2 (1)+Li_2 ((1/2))−(3/2)Li_2 (−1)) ∫_0 ^1 ((ln(1+x^2 ))/(1+x))dx=ln^2 (2)−A =((ln^2 (2))/2)+(π^2 /(48))−((Li_2 ((1/2)))/2)=(3/4)ln^2 (2)−(π^2 /(48)) Li_2 ((1/2))=(π^2 /(12))−((ln^2 (2))/2)](https://www.tinkutara.com/question/Q206584.png)
$$=\left[{ln}\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−{A} \\ $$$${B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\mathrm{2}{x},{A}+{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{xln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{2}\right)+{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}}\right)}{dx} \\ $$$$\left.=\left.{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}}\right)\right.}{\frac{\mathrm{1}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}}}{d}\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−{Li}_{\mathrm{2}} \left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${B}−{A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{xln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx};{t}=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}\left(\mathrm{1}−{t}\right)}{\mathrm{1}+{t}}.\frac{{ln}\left({t}\right)}{\frac{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }}.\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{t}\right)\left({ln}\left({t}\right)\right.}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\mathrm{1}+{t}}+\frac{−{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{ln}\left({t}\right)=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{{t}}{dt}_{{t}^{\mathrm{2}} \rightarrow{t}} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}{dt}=−\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\left(−{t}\right)\right)}{\left(−{t}\right)}{d}\left(−{t}\right)= \\ $$$$\left.\frac{\mathrm{3}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(−{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{3}{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({B}+{A}\right)−\left({B}−{A}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}^{\mathrm{2}} \left(\mathrm{2}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{3}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}}{dx}={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−{A} \\ $$$$=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{48}}−\frac{{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$${Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}} \\ $$
Commented by York12 last updated on 19/Apr/24
$$ \\ $$$$\mathrm{Perfect}\:! \\ $$$$\mathrm{but}\:\mathrm{there}\:\mathrm{is}\:\mathrm{probably}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{somewhere} \\ $$$$\mathrm{cause}\:\mathrm{the}\:\mathrm{final}\:\mathrm{answer}\:\mathrm{must}\:\mathrm{be}\:\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}\:^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$
Commented by Berbere last updated on 19/Apr/24
$${yes}\:\:{you}\:{right}\:{i}\:{corrected}\:{it} \\ $$
Commented by York12 last updated on 19/Apr/24
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Berbere last updated on 19/Apr/24
$${withe}\:{Pleasur}\: \\ $$