-1-1- e x dx is equal toA- log eex-1-ex-CB- log eex-ex-1-CC- log eex-1-ex-CD- log eex-ex-1-C

The correct option is B log_e ( e^x/e^x+1)+C nbsp; nbsp;Let 1+e^x=t ⇒ e^x dx =dt ⇒ nbsp; dx =dt/t 1 ∫1/1+e^xdx= ∫dt/t(t 1)=∫(1/t 1 1/t)dt = log (t 1) log ...

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Byju's Answer

Standard XII

Mathematics

Property 1

∫1/1+ e x dx ...

Question

$\int \frac{1}{1 + e^{x}} d x$ is equal to

{log}_{e} (\frac{e^{x} + 1}{e^{x}}) + C

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{log}_{e} (\frac{e^{x} - 1}{e^{x}}) + C

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{log}_{e} (\frac{e^{x}}{e^{x} + 1}) + C

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{log}_{e} (\frac{e^{x}}{e^{x} - 1}) + C

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Solution

The correct option is C ${log}_{e} (\frac{e^{x}}{e^{x} + 1}) + C$
$Let 1 + e^{x} = t$
$\Rightarrow e^{x} d x = d t$
$\Rightarrow d x = \frac{d t}{t - 1}$
$\int \frac{1}{1 + e^{x}} d x = \int \frac{d t}{t (t - 1)} = \int (\frac{1}{t - 1} - \frac{1}{t}) d t$
$= log (t - 1) - log (t) + C$
$= log (\frac{t - 1}{t}) + C$
$= log (\frac{e^{x}}{e^{x} + 1}) + C$

Suggest Corrections

∫11+ex dx is equal to

The correct option is C loge(exex+1)+C Let 1+ex=t ⇒exdx=dt ⇒dx=dtt−1 ∫11+exdx=∫dtt(t−1)=∫(1t−1−1t)dt =log(t−1)−log(t)+C =log(t−1t)+C =log(exex+1)+C

$\int \frac{1}{1 + e^{x}} d x$ is equal to