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Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 22

Answers (1)

Answer:

        -\frac{a}{b}log\left | be^{-x}+c \right |+C

Hint:

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

Given:

        \int\! \! \frac{a}{b+ce^{x}}dx

Explanation:

        \int\! \! \frac{a}{e^{x}(\frac{b}{e^{x}}+c)}dx            .......(1)

        =\int\! \! \frac{ae^{-x}}{be^{-x}+c}dx

Let

        be^{-x}+c=t

        -e^{-x}(b)dx=dt

        e^{-x}dx=-\frac{1}{b}dt

Put in (1)

        \int \! \frac{a (-\frac{1}{b})dt}{t}

        =-\frac{a}{b}\int \! \frac{dt}{t}

        =-\frac{a}{b}log\left | t \right |+C

        =-\frac{a}{b}log\left | be^{-x}+c \right |+C

Posted by

Gurleen Kaur

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