Question #daa0e

Dec 29, 2017

$\int \frac{1}{a + b {e}^{x}} \mathrm{dx} = - \frac{\ln | a {e}^{- x} + b |}{a} + C = \frac{x}{a} - \frac{\ln | a + b {e}^{x} |}{a} + C$

Explanation:

Let $f \left(x\right) = \frac{1}{a + b {e}^{x}}$ and note that we can multiply the numerator (top) and denominator (bottom) of this fraction by ${e}^{- x}$ to write $f \left(x\right) = \frac{{e}^{- x}}{a {e}^{- x} + b}$.

For the integral $\int \frac{{e}^{- x}}{a {e}^{- x} + b} \mathrm{dx}$, use the substitution $u = a {e}^{- x} + b$, $\mathrm{du} = - a {e}^{- x} \mathrm{dx}$, and ${e}^{- x} \mathrm{dx} = - \frac{1}{a} \mathrm{du}$ to obtain

$\int \frac{{e}^{- x}}{a {e}^{- x} + b} \mathrm{dx} = - \frac{1}{a} \int \setminus \frac{1}{u} \mathrm{du}$

$= - \frac{1}{a} \ln | u | + C = - \frac{\ln | a {e}^{- x} + b |}{a} + C$.

To see that this is the same as the second answer, note that
$x = \ln \left({e}^{x}\right) = \ln | {e}^{x} |$ so that, by properties of logarithms,

$x - \ln | a + b {e}^{x} | = \ln | {e}^{x} | - \ln | a + b {e}^{x} |$

$= \ln | \frac{{e}^{x}}{a + b {e}^{x}} | = - \ln | \frac{a + b {e}^{x}}{{e}^{x}} | = - \ln | a {e}^{- x} + b |$.

Therefore, $\int \frac{1}{a + b {e}^{x}} \mathrm{dx} = - \frac{\ln | a {e}^{- x} + b |}{a} + C = \frac{x}{a} - \frac{\ln | a + b {e}^{x} |}{a} + C$.