Menu Close

0-1-ln-2-ln-1-x-2-1-x-dx-

Tinku Tara 0 Comments
Pin




Question Number 140866 by liberty last updated on 13/May/21
∫_0 ^1 ((ln 2−ln (1+x^2 ))/(1−x)) dx =?
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{x}}\:\mathrm{dx}\:=?\: \\ $$
Answered by mindispower last updated on 15/May/21
=lim_(x→1) ∫_0 ^x ((ln(2)−ln(1+t^2 ))/(1−t))dt =lim_(x→1) −ln(2)ln(1−x)+ln(1−x)ln(1+x^2 )−∫_0 ^x ((ln(1−t)2t)/(1+t^2 ))dt =lim_(x→1) ln(1−x).ln(((1+x^2 )/2))=lim_(x→1) ln(1−((1−x^2 )/2))ln(1−x)=0 =−∫_0 ^1 ((ln(1−x).2xdx)/(1+x^2 ))=−A let B=∫_0 ^1 ((ln(1+x).2xdx)/(1+x^2 )) A+B=∫_0 ^1 ((ln(1−x^2 ))/(1+x^2 )).d(x^2 ) =∫_0 ^1 ((ln(1−t))/(1+t))dt=∫_0 ^1 ((ln(t))/(2−t))dt =∫_0 ^(1/2) ((ln(2w))/(1−w)).dw=ln^2 (2)+∫_0 ^(1/2) ((ln(w))/(1−w))dw =ln^2 (2)+∫_1 ^(1/2) −((ln(1−t))/t)=ln^2 (2)+li_2 ((1/2))−li_2 (1) A−B=∫_0 ^1 ((ln(((1−t)/(1+t))).2tdt)/(1+t^2 )),t=((1−x)/(1+x)) dt=((−2)/((1+x)^2 ))dx =4∫_0 ^1 ((ln(x))/(2+2x^2 )).((1−x)/(1+x))dx ∫_0 ^1 (((1−x)ln(x))/((1+x)(1+x^2 )))dx =∫_0 ^1 ((ln(x))/(1+x))−(x/(1+x^2 ))ln(x)dx =(3/4)∫_0 ^1 ((ln(x))/(1+x))dx=(3/4).−∫_0 ^1 ((ln(1−(−x)))/(−x)).d(−x) =−∫_0 ^(−1) ((ln(1−t))/t)dt=((−3)/4)li_2 (−1)
$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\int_{\mathrm{0}} ^{{x}} \frac{{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}−{t}}{dt} \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}−{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{1}−{x}\right)+{ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\int_{\mathrm{0}} ^{{x}} \frac{{ln}\left(\mathrm{1}−{t}\right)\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{ln}\left(\mathrm{1}−{x}\right).{ln}\left(\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}}\right)=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{ln}\left(\mathrm{1}−\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}}\right){ln}\left(\mathrm{1}−{x}\right)=\mathrm{0} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right).\mathrm{2}{xdx}}{\mathrm{1}+{x}^{\mathrm{2}} }=−{A} \\ $$$${let}\:{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right).\mathrm{2}{xdx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${A}+{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }.{d}\left({x}^{\mathrm{2}} \right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}+{t}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\mathrm{2}−{t}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}\left(\mathrm{2}{w}\right)}{\mathrm{1}−{w}}.{dw}={ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}\left({w}\right)}{\mathrm{1}−{w}}{dw} \\ $$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}={ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{li}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$${A}−{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\right).\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{2}} },{t}=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}} \\ $$$${dt}=\frac{−\mathrm{2}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} }.\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right){ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{ln}\left({x}\right){dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}{dx}=\frac{\mathrm{3}}{\mathrm{4}}.−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\left(−{x}\right)\right)}{−{x}}.{d}\left(−{x}\right) \\ $$$$=−\int_{\mathrm{0}} ^{−\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}=\frac{−\mathrm{3}}{\mathrm{4}}{li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *