Q56 / 150 · MHT CET 2019 (Shift 1) · Mathematics

${\int l o g x . [\log (e x)]}^{- 2} d x = \dots$

Correct answer
A
$\frac{x}{1 + l o g x} + c$
B
$x (1 - l o g x) + c$
C
$x (1 + l o g x) + c$
D
$\frac{x}{1 - l o g x} + c$

Solution

Let

l = \int l o g \times {[\log (e x)]}^{- 2} d x

= \int \frac{l o g x}{{(l o g e x)}^{2}} d x

= \int \frac{l o g x}{{(l o g e + l o g x)}^{2}} d x

= \int \frac{l o g x}{{(1 + l o g x)}^{2}} d x

P u t l o g x = t \Rightarrow x = e^{t}

\Rightarrow d x = e^{t} d t

= \int \frac{t}{{(1 + t)}^{2}} e^{t} d t

= \int e^{t} (\frac{1}{1 + t} - \frac{1}{{(1 + t)}^{2}}) d t

= \frac{e^{t}}{1 + t} + C

= \frac{x}{(1 + l o g x)} + C

Hence, the correct answer is option: 'A'.

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