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  • Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 23

Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 23

Answers (1)

Answer:

        -log\left | e^{-x}+1 \right |+C

Hint:

        \int \! \frac{1}{x}dx=log\left | x \right |+C

Given:

        \int \! \frac{1}{e^{x}+1}dx

Explanation:

        \int \! \frac{1}{e^{x}(1+\frac{1}{e^{x}})}dx\; \; \; \; =\int \! \frac{e^{-x}}{e^{-x}+1}dx                ....(1)

Let

        e^{-x}+1=t

        e^{-x}dx=-dt

From (1)

        \int \! \frac{-dt}{t}=-log\left | t \right |+C

        =-log\left | e^{-x}+1 \right |+C

Posted by

Gurleen Kaur

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