integral of e^(-x)/(9e^(-2x)+1)^(3/2)

Before integrating, let's calculate:

\frac{e^{- x}}{{(9 e^{(-2 x)} + 1)}^{\frac{3}{2}}}

\frac{\frac{1}{e^{x}}}{{(9 e^{(- 2 x)} + 1)}^{\frac{3}{2}}}

\frac{1}{e^{x}} \cdot \frac{1}{{(\frac{9}{e^{(2 x)}} + 1)}^{\frac{3}{2}}}

\frac{1}{e^{x} \cdot {(\frac{9}{e^{(2 x)}} + 1)}^{\frac{3}{2}}}

Let's compute the integral:

\int(\frac{1}{e^{x} \cdot {(\frac{9}{e^{(2 x)}} + 1)}^{\frac{3}{2}}})dx

We look for an antiderivative of the form

\frac{\frac{1}{e^{x}}}{\sqrt{\frac{9}{e^{(2 x)}} + 1}}

. The quotient rule and the derivative of the square root give

{(\frac{\frac{1}{e^{x}}}{\sqrt{\frac{9}{e^{(2 x)}} + 1}})}^{'} = - \frac{e^{x}}{\sqrt{\frac{9}{e^{(2 x)}} + 1} \cdot (9 + e^{(2 x)})}

The integrand is

-1

times this derivative, so we multiply the antiderivative by

-1

This calculation is valid on each real interval where

\frac{9}{e^{(2 x)}} + 1 > 0

- \frac{1}{e^{x} \cdot \sqrt{\frac{9}{e^{(2 x)}} + 1}}

▶Indefinite integrals are defined up to an additive constant C, so this yields:

- \frac{1}{e^{x} \cdot \sqrt{\frac{9}{e^{(2 x)}} + 1}} + C