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Question
Evaluate the following integrals : `int_0^1 log(1/x - 1)*dx`
Solution 1
Let I = `int_0^1 log(1/x - 1)*dx`
∴ I = `int_0^1 log((1 - x)/x)*dx` ...(i)
= `int_0^1 log[(1 - (1 - x))/(1 - x)]*dx ...[because int_0^"a" f(x)*dx = int_0^"a" f("a" - x)*dx]`
I = `int_0^"a" log(x/(1 - x))*dx` ...(ii)
Adding (i) and (ii), we get
2I = `int_0^1 log((1 - x)/x)*dx + int_0^1 log(x/(1 - x))*dx`
= `int_0^1[log ((1 - x)/x) + log (x/(1 - x))]*dx`
= `int_0^1 log ((1 - x)/x xx x/(1 - x))*dx`
= `int_0^1 log 1*dx`
∴ 2I = `int_0^1 0*dx`
∴ I = 0.
Solution 2
Let I = `int_0^1 log(1/x - 1)*dx`
= `int_0^1 log((1 - x)/x)*dx`
= `int_0^1 [log(1 - x) - logx]*dx` ...(1)
We use the property `int_0^a f(x)*dx = int_0^a f(a - x)*dx`
Here, a = 1
Hence in I, changing x to 1 – x, we get
I = `int_0^1 [log |1 - (1 - x)| - log(1 - x)]*dx`
= `int_0^1 [logx - log(1 - x)]*dx`
= `-int_0^1 [log (1 - x) - logx]*dx`
= – 1 ...[By (1)]
∴ 2I = 0
∴ I = 0.
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