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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 16

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Answer:-\log|e^{-x}+2|+c

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

Given: \int \frac{1}{1+2e^{x}}dx

Solution: \int \frac{1}{1+2e^{x}}dx

=\int \frac{e^{-x}}{e^{-x}(1+2e^{x})}dx

=\int \frac{e^{-x}}{e^{-x}+2}dx                          [ multiply and divide by e^{x}]

Lett = e^{-x}+2   and differentiate both sides, \left [ \frac{d}{dx}e^{-x}= -e^{-x} \right ]

 dt = -e^{-x} dx

I=\int \frac{-dt}{t}

=-\log \left | t \right |+c

=-\log \left | e^{-x} +2\right |+c                                                                  \left [\int \frac{1}{x}dx=\log x+c \right ]

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