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Evaluate the following definite integral :

$\int \limits_0^1 \log \left(\frac{1}{x} - 1 \right)$
asked in Calculus by Veteran (24.5k points)   | 200 views
Apply integration by parts,

it will be evaluated to 0.
Yes, answer is 0, but i didn't get it.

Let me try once more.
by parts we r getting $x log(1/x-1)+log(1-x) =0$

Your 2nd term is wrong.

it is -ln(1 - x)

why r u trying with ln, try with simple log
still no difference
yes, my mistake :(

2 Answers

+2 votes
Best answer
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.
answered by Junior (617 points)  
selected by
which property it is ?
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
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 by Junior (625 points)  


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