The solution of $(e^{x} + 1) y d y = (y + 1) e^{x} d x$ is

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

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

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$e^{y} = C [(y + 1) (e^{x} + 1)]$

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

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Solution

The correct option is A
$e^{y} = C [(y + 1) (e^{x} + 1)]$

Consider the given integral.
$(e^{x} + 1) y d y = (y + 1) e^{x} d x$

So,
$\frac{y d y}{y + 1} = \frac{e^{x}}{(e^{x} + 1)} d x$
$(1 - \frac{1}{y + 1}) d y = \frac{e^{x}}{(e^{x} + 1)} d x$

Integrate both sides.
$\int (1 - \frac{1}{y + 1}) d y = \int \frac{e^{x}}{(e^{x} + 1)} d x$

Let,
$e^{x} + 1 = u$
$e^{x} d x = d u$

Therefore,
$\int (1 - \frac{1}{y + 1}) d y = \int \frac{e^{x}}{(e^{x} + 1)} d x$
$y - ln (y + 1) = \int \frac{d u}{d}$
$y - ln (y + 1) = ln u + c$
$y - ln (y + 1) = ln (e^{x} + 1) + ln C$
$y = ln (y + 1) + ln (e^{x} + 1) + ln C$
$y = ln [(y + 1) \times (e^{x} + 1) \times C]$

Take anti-log on both the sides.
$e^{y} = C [(y + 1) (e^{x} + 1)]$

Hence, this is the required result.

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