Question Number 502 by 123456 last updated on 25/Jan/15
$$\underset{−\mathrm{1}} {\overset{+\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\mid{x}\mid}{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$
Answered by prakash jain last updated on 20/Jan/15
![I=∫_(−1) ^(+1) ((ln ∣x∣)/(x^2 −1))dx=2∫_0 ^1 (1/2){((ln x)/(x−1))−((ln x)/(x+1))}dx .....(i) ∫_0 ^1 ((ln x)/(x+1))dt=∫_1 ^0 ((ln (1−t))/(−t))(−dt) , x=1−t, dx=−dt =∫_1 ^0 (1/t)[−t−(t^2 /2)−(t^3 /3)−(t^4 /4)−..]dt =∫_1 ^0 [−1−(t/2)−(t^2 /3)−(t^3 /4)−...]dt =[1+(1/2^2 )+(1/3^2 )+(1/4^2 )+(1/5^2 )+...]=(π^2 /6) Similarly ∫_0 ^1 ((ln x)/(x+1))=−(π^2 /(12)) Putting values in (i) I={(π^2 /6) −(−(π^2 /(12)))}=(π^2 /4)](https://www.tinkutara.com/question/Q504.png)
$${I}=\underset{−\mathrm{1}} {\overset{+\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\mid{x}\mid}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{ln}\:{x}}{{x}−\mathrm{1}}−\frac{\mathrm{ln}\:{x}}{{x}+\mathrm{1}}\right\}{dx}\:\:\:\:\:…..\left(\mathrm{i}\right) \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:{x}}{{x}+\mathrm{1}}{dt}=\overset{\mathrm{0}} {\int}_{\mathrm{1}} \frac{\mathrm{ln}\:\left(\mathrm{1}−{t}\right)}{−{t}}\left(−{dt}\right)\:,\:{x}=\mathrm{1}−{t},\:{dx}=−{dt} \\ $$$$=\underset{\mathrm{1}} {\overset{\mathrm{0}} {\int}}\frac{\mathrm{1}}{{t}}\left[−{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{{t}^{\mathrm{4}} }{\mathrm{4}}−..\right]{dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{0}} \:\left[−\mathrm{1}−\frac{{t}}{\mathrm{2}}−\frac{{t}^{\mathrm{2}} }{\mathrm{3}}−\frac{{t}^{\mathrm{3}} }{\mathrm{4}}−…\right]{dt} \\ $$$$=\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+…\right]=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{Similarly} \\ $$$$\overset{\mathrm{1}} {\int}_{\mathrm{0}} \frac{\mathrm{ln}\:{x}}{{x}+\mathrm{1}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\mathrm{Putting}\:\mathrm{values}\:\mathrm{in}\:\left(\mathrm{i}\right) \\ $$$${I}=\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)\right\}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$