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2 ∫ 1 1 X ( 1 + Log X ) 2 D X

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Question

\[\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx\]
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Solution

\[Let\ 1 + \log\ x\ = t . Then, \frac{1}{x}\ dx\ = dt\]
\[When\ x = 1, t = 1\ and\ x\ = 2, t = \left( 1 + \log 2 \right)\]
\[ \therefore I = \int_1^2 \frac{1}{x \left( 1 + \log x \right)^2} d x\]
\[ \Rightarrow I = \int_1^\left( 1 + \log 2 \right) \frac{1}{t^2} dt\]
\[ \Rightarrow I = \left[ \frac{- 1}{t} \right]_1^\left( 1 + \log 2 \right) \]
\[ \Rightarrow I = - \frac{1}{\left( 1 + \log 2 \right)} + 1\]
\[ \Rightarrow I = \frac{\log 2}{\log\ 2 + \log e}\]
\[ \Rightarrow I = \frac{\log 2}{\log\ 2e}\]

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Chapter 19: Definite Integrals - Exercise 20.2 [Page 38]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.2 | Q 2 | Page 38

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