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Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 12 maths

Answers (1)

Answer:

        \frac{1}{3}log\left | e^{3x}+1 \right |+C

Hint:

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

Given:

        \int \! \frac{e^{3x}}{e^{3x}+1}dx

Explanation:

{e^{3x}+1}=t

3e^{3x}dx=dt

            =\frac{1}{3}\! \int \! \frac{dt}{t}

            =\frac{1}{3}log\left | t \right |

            =\frac{1}{3}log\left | e^{3x} \right |+C

Posted by

Gurleen Kaur

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