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प्रश्न
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
बेरीज
उत्तर
Let I = `int_0^1 log (1/x - 1) "d"x`
I = `int_0^1 log ((1 - x)/x) "d"x` ........(1)
Using the property
`int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" -x) "d"x`
I = `int_0^1 log (1/(1 - x) - 1) "d"x`
= `int_0^1 log((1 - (1 - x))/(1 - x)) "d"x`
= `int_0^1 log(x/(1 - x)) "d"x`
Adding (1) and (2)
I + I = `int_0^1 log((1 - x)/x) "d"x + int_0^1 log (x/(1 - x )) "d"x`
2I = `int_0^1 log [((1 - x))/x xx x/((1 - x))] "d"x`
2I = `int_0^1 log(1) "d"x` = 0
∴ I = 0
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Definite Integrals
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