Question #daa0e

1 Answer
Dec 29, 2017

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.