$\int \frac{a}{b + c e^{x}} d x$ is equal to

\frac{a}{b} log | b + c \cdot e^{x} | + K

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- \frac{a}{b} log | c + b e^{- x} | + K

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log | b + c \cdot e^{x} | + K

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log | c + b \cdot e^{- x} | + K

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Solution

The correct option is B $- \frac{a}{b} log | c + b e^{- x} | + K$
multiplying numerator and denominator with $e^{- x}$
$\int \frac{a e^{- x}}{b e^{- x} + c} d x$
Let $t = b e^{- x} + c, d t = - b e^{- x} d x$
Then
$I = \frac{- a}{b} \int \frac{d t}{t}$

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