1 votes 1 votes Evaluate the following definite integral : $\int \limits_0^1 \log \left(\frac{1}{x} - 1 \right)$ Calculus integration calculus engineering-mathematics integrals + – mcjoshi asked Sep 4, 2016 mcjoshi 1.2k views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Kapil commented Sep 4, 2016 reply Follow Share Apply integration by parts, it will be evaluated to 0. 0 votes 0 votes mcjoshi commented Sep 4, 2016 reply Follow Share Yes, answer is 0, but i didn't get it. Let me try once more. 1 votes 1 votes srestha commented Sep 4, 2016 i edited by srestha Sep 4, 2016 reply Follow Share by parts we r getting $x log(1/x-1)+log(1-x) =0$ 0 votes 0 votes Kapil commented Sep 4, 2016 reply Follow Share Your 2nd term is wrong. it is -ln(1 - x) 0 votes 0 votes srestha commented Sep 4, 2016 reply Follow Share why r u trying with ln, try with simple log 0 votes 0 votes Kapil commented Sep 4, 2016 reply Follow Share still no difference 0 votes 0 votes srestha commented Sep 4, 2016 reply Follow Share yes, my mistake :( 0 votes 0 votes Please log in or register to add a comment.
Best answer 3 votes 3 votes Applying the property of logarithms: $log(\frac{1}{x}-1) = log(\frac{1-x}{x}) = log(1-x) - log(x)$ Integrating from 0 to 1 we can see that the same values are repeated in each term thus reducing the answer to zero. krish__ answered Sep 10, 2016 selected Feb 20, 2017 by mcjoshi krish__ comment Share Follow See all 2 Comments See all 2 2 Comments reply mcjoshi commented Sep 10, 2016 reply Follow Share which property it is ? 0 votes 0 votes krish__ commented Sep 10, 2016 reply Follow Share Since log is a monotonically increasing function: $\int_{0}^{1} log(x) = \int_{0}^{1}log(1-x)$ and we know that log(a/b) = log(a) - log(b) 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes I = 0 Prateek kumar answered Jan 4, 2018 Prateek kumar comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes ans is 0 still i want to cnfirm xlg((1/x)-1)-log(1-x) on putting x=0 ///// lg((1/x)-1) this will be undefined __ answered Sep 4, 2016 __ comment Share Follow See all 0 reply Please log in or register to add a comment.