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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 15

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Answer:-\log|1+e^{-x}|+c

Hint: You must know about the integration rule of exponential function.

Given= \int \frac{1}{1+e^{x}}dx

Solution:

\begin{aligned} &\int \frac{1}{1+e^{x}} d x \\ &=\int \frac{e^{-x}}{e^{-x}\left(1+e^{x}\right)} d x \quad\quad\quad\quad\quad\quad\quad\quad\left [ multiply\: and \: divide\: by \: e^{x} \right ]\\ &=\int \frac{e^{-x}}{1+e^{-x}} d x \end{aligned}

Let e^{-x}+1=t                                                                              \left [ \frac{d}{dx}e^{-x}= -e^{-x} \right ]

\begin{aligned} &-\mathrm{e}^{-x} \mathrm{~d} \mathrm{x}=\mathrm{dt} \\ &I=\int \frac{-d t}{t} \\ &=-\log |t|+c \\ &=-\log \left|e^{-x}+1\right|+c\quad\quad\quad\quad\quad\quad \quad\left[\int \frac{1}{x} d x=\log x+c\right] \end{aligned}

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