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NIMCET 2017 Question Paper with solutions

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Question : 113 of 120

Marks: +1, -0

The solution of

(e^x + 1)

e^x

dx is

$e^y$ = c ( $e^x$ + 1) (y +1)
$e^y$ = $e^x$ + y + 1
y = ( $e^x$ +1) (y + 1)
None of these

Solution:

Concept:

\log x=y \Rightarrow x=e^{y}

Calculation:

Given:

(e^{x}+1) y d y=(y+1) e^{x} d x

\Rightarrow \;\frac{y dy}{y+1} = \;\frac{e^{x}}{e^{x}+1} dx

\Rightarrow \;\frac{(y+1-1) dy}{y+1} = \;\frac{e^{x}}{e^{x}+1} dx

\Rightarrow dy - \;\frac{dy}{y+1} = \;\frac{e^{x}}{e^{x}+1} dx

Integrating both sides, we get

\Rightarrow y - \log |y+1| = \log (e^{x}+1) + \log c

\Rightarrow y = \log |(y+1)(e^{x}+1)| + \log c

\Rightarrow c(y+1)(e^{x}+1) = e^{y}

\therefore e^{y} = c(e^{x}+1)(y+1)

Hence, option (1) is correct.

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