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Evaluate the following problems using properties of integration: $\int\limits_{0}^{1}\log\large(\frac{1}{x}$$-1)$$dx$

1 Answer

Toolbox:
  • $\int \limits_0^a f(x) dx=\int \limits_0^a f(a-x) dx$
$\int\limits_{0}^{1}\log\large(\frac{1}{x}$$-1)$$dx$
Step1:
$\int\limits_{0}^{1}\log\large(\frac{1}{x}$$-1)$$dx=I$
Step 2:
$\qquad= \int\limits_{0}^{1}\log\large(\frac{1}{1-x}$$-1)$$dx$
$\qquad=\int\limits_{0}^{1}\log\large\frac{1-(1-x)}{1-x}$$dx$
$\qquad=\int\limits_{0}^{1}\log\large\frac{x}{1-x}$$dx$
Step 3:
$2I= \int\limits_{0}^{1}\bigg[\log\large(\frac{1}{x}$$-1)$$+\log\large(\frac{x}{1-x})\bigg]$$dx$
$\quad= \int\limits_{0}^{1}\bigg[\log\large(\frac{1-x}{x})$$+\log\large(\frac{x}{1-x})\bigg]$$dx$
$\quad= \int\limits_{0}^{1}\log\large\bigg (\frac{1-x}{x}.\frac{x}{1-x}\bigg)$$dx$
$\quad=\int \limits _0^1 \log \; 1 \;dx$$=0$
$\quad=I=0$
answered Aug 14, 2013 by meena.p
 

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