Question

Question #daa0e

Answer 1

`int 1/(a+be^(x))dx=-(ln|ae^{-x}+b|)/a+C=x/a-(ln|a+be^{x}|)/a+C`

Explanation:

Let `f(x)=1/(a+be^(x))` and note that we can multiply the numerator (top) and denominator (bottom) of this fraction by `e^(-x)` to write `f(x)=(e^{-x})/(ae^{-x}+b)`.

For the integral `int (e^{-x})/(ae^{-x}+b)dx`, use the substitution `u=ae^{-x}+b`, `du=-ae^{-x}dx`, and `e^{-x}dx=-1/a du` to obtain

`int (e^{-x})/(ae^{-x}+b)dx=-1/a int\ 1/u du`

`=-1/a ln|u|+C=-(ln|ae^{-x}+b|)/a+C`.

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

`x-ln|a+be^{x}|=ln|e^{x}|-ln|a+be^{x}|`

`=ln|(e^{x})/(a+be^{x})|=-ln|(a+be^{x})/(e^{x})|=-ln|ae^{-x}+b|`.

Therefore, `int 1/(a+be^(x))dx=-(ln|ae^{-x}+b|)/a+C=x/a-(ln|a+be^{x}|)/a+C`.