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Question Bank [2022] — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Question Bank [2022]3 Marks
Question
Evaluate the following integrals: $\int_0^1 \log \left(\frac{1}{x}-1\right) \cdot d x$
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Answer

$\text { Let } I =\int_0^1 \log \left(\frac{1}{x}-1\right) \cdot d x$
$\therefore I =\int_0^1 \log \left(\frac{1-x}{x}\right) \cdot d x \quad \ldots \text { (i) }$
$=\int_0^1 \log \left[\frac{1-(1-x)}{1-x}\right] \cdot d x \quad \ldots\left[\because \int_0^{ a } f(x) \cdot d x=\int_0^{ a } f( a -x) \cdot d x\right]$
$I =\int_0^{ a } \log \left(\frac{x}{1-x}\right) \cdot d x \quad \ldots \text { (ii) }$
Adding (i) and (ii), we get
$ \text { 2। }=\int_0^1 \log \left(\frac{1-x}{x}\right) \cdot d x+\int_0^1 \log \left(\frac{x}{1-x}\right) \cdot d x$
$=\int_0^1\left[\log \left(\frac{1-x}{x}\right)+\log \left(\frac{x}{1-x}\right)\right] \cdot d x$
$=\int_0^1 \log \left(\frac{1-x}{x} \times \frac{x}{1-x}\right) \cdot d x$
$=\int_0^1 \log 1 \cdot d x$
$\therefore 2 I =\int_0^1 0 \cdot d x$
$\therefore I =0 . $

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