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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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