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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Find $ \int \left[ \log ( \log x ) + \frac { 1 } { ( \log x ) ^ { 2 } } \right] d x$
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Answer

According to the question ,  $ I = \int \left[ \log ( \log x ) + \frac { 1 } { ( \log x ) ^ { 2 } } \right] d x$
By Using integration by parts for first integral, we get
$ = \int {\log } (\mathop {\mathop{\rm l}\nolimits} \limits_Iogx) \cdot \mathop 1\limits_{II} dx + \int {\frac{1}{{{{(\log x)}^2}}}} dx$
$I = log (log x) \int 1 d x - \int \left[ \frac { d } { d x } \log ( \log x )\Big]\Big[ \int 1 d x \right] d x + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x$

$= x log (log x) - \int \frac { 1 } { ( \log x ) } d x + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $

$= x log (log x)-\int (\mathop {\mathop{\rm\frac { 1}{logx}}\nolimits} \limits_I ) \cdot \mathop 1\limits_{II} dx + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $

By Using integration by parts for second integral, we get

$= x log (log x) -\frac{x}{logx}+ \int [\frac {d}{dx}(\frac {1}{logx})]. (\int 1dx)dx+ \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$= x log (log x) -\frac{x}{logx}- \int \frac { 1 } { ( \log x ) ^ { 2 } } d x \frac {1}{x} . x+ \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $

$= x log (log x) -\frac{x}{logx}- \int \frac { 1 } { ( \log x ) ^ { 2 } } d x + \int \frac { 1 } { ( \log x ) ^ { 2 } } d x $
$ = x \log ( \log x )-\frac{x}{logx}+ C$
$\therefore I = x \log ( \log x ) -\frac{x}{logx}+ C$

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