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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : $ \int_0^1 \log \bigg(\large\frac{1}{x}-1\bigg)$$dx $

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Toolbox:
  • $\int\limits_0^af(x)dx=\int_0^af(a-x)dx$
Step 1:
$I=\int_0^1\log(\large\frac{1}{x}$$-1)dx$
$\;\;=\int_0^1\log\big(\large\frac{1-x}{x}\big)$$dx$------(1)
Step 2:
$\int\limits_0^af(x)dx=\int_0^af(a-x)dx$
$I=\int_0^1\log\bigg[\large\frac{1-(1-x)}{1-x}\bigg]$$dx$
$I=\int\limits_0^1\log\bigg[\large\frac{1-(1-x)}{1-x}\bigg]$$dx$
$I=\int_0^1\log\big(\large\frac{x}{1-x}\big)$$dx$-----(2)
Step 3:
Add (1) and (2) we get
$2I=\int_0^1\log\big(\large\frac{1-x}{x}\big)+$$\log\big(\large\frac{x}{1-x}\big)$$dx$
$\;\;=\int_0^1\log(1)=0$
answered Nov 14, 2013 by sreemathi.v
 
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