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प्रश्न
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
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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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संबंधित प्रश्न
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Evaluate the following:
Γ(4)
