Integral of $$$\frac{e^{- x}}{16 - 9 e^{- 2 x}}$$$

Find $$$\int \frac{e^{- x}}{16 - 9 e^{- 2 x}}\, dx$$$.

Simplify:

$${\color{red}{\int{\frac{e^{- x}}{16 - 9 e^{- 2 x}} d x}}} = {\color{red}{\int{\frac{e^{x}}{16 e^{2 x} - 9} d x}}}$$

Let $$$u=4 e^{x}$$$.

Then $$$du=\left(4 e^{x}\right)^{\prime }dx = 4 e^{x} dx$$$ (steps can be seen »), and we have that $$$e^{x} dx = \frac{du}{4}$$$.

So,

$${\color{red}{\int{\frac{e^{x}}{16 e^{2 x} - 9} d x}}} = {\color{red}{\int{\frac{1}{4 \left(u^{2} - 9\right)} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \frac{1}{u^{2} - 9}$$$:

$${\color{red}{\int{\frac{1}{4 \left(u^{2} - 9\right)} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u^{2} - 9} d u}}{4}\right)}}$$

Perform partial fraction decomposition (steps can be seen »):

$$\frac{{\color{red}{\int{\frac{1}{u^{2} - 9} d u}}}}{4} = \frac{{\color{red}{\int{\left(- \frac{1}{6 \left(u + 3\right)} + \frac{1}{6 \left(u - 3\right)}\right)d u}}}}{4}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(- \frac{1}{6 \left(u + 3\right)} + \frac{1}{6 \left(u - 3\right)}\right)d u}}}}{4} = \frac{{\color{red}{\left(\int{\frac{1}{6 \left(u - 3\right)} d u} - \int{\frac{1}{6 \left(u + 3\right)} d u}\right)}}}{4}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(u \right)} = \frac{1}{u + 3}$$$:

$$\frac{\int{\frac{1}{6 \left(u - 3\right)} d u}}{4} - \frac{{\color{red}{\int{\frac{1}{6 \left(u + 3\right)} d u}}}}{4} = \frac{\int{\frac{1}{6 \left(u - 3\right)} d u}}{4} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{u + 3} d u}}{6}\right)}}}{4}$$

Let $$$v=u + 3$$$.

Then $$$dv=\left(u + 3\right)^{\prime }du = 1 du$$$ (steps can be seen »), and we have that $$$du = dv$$$.

Therefore,

$$\frac{\int{\frac{1}{6 \left(u - 3\right)} d u}}{4} - \frac{{\color{red}{\int{\frac{1}{u + 3} d u}}}}{24} = \frac{\int{\frac{1}{6 \left(u - 3\right)} d u}}{4} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{24}$$

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$\frac{\int{\frac{1}{6 \left(u - 3\right)} d u}}{4} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{24} = \frac{\int{\frac{1}{6 \left(u - 3\right)} d u}}{4} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{24}$$

Recall that $$$v=u + 3$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{24} + \frac{\int{\frac{1}{6 \left(u - 3\right)} d u}}{4} = - \frac{\ln{\left(\left|{{\color{red}{\left(u + 3\right)}}}\right| \right)}}{24} + \frac{\int{\frac{1}{6 \left(u - 3\right)} d u}}{4}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(u \right)} = \frac{1}{u - 3}$$$:

$$- \frac{\ln{\left(\left|{u + 3}\right| \right)}}{24} + \frac{{\color{red}{\int{\frac{1}{6 \left(u - 3\right)} d u}}}}{4} = - \frac{\ln{\left(\left|{u + 3}\right| \right)}}{24} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{u - 3} d u}}{6}\right)}}}{4}$$

Let $$$v=u - 3$$$.

Then $$$dv=\left(u - 3\right)^{\prime }du = 1 du$$$ (steps can be seen »), and we have that $$$du = dv$$$.

So,

$$- \frac{\ln{\left(\left|{u + 3}\right| \right)}}{24} + \frac{{\color{red}{\int{\frac{1}{u - 3} d u}}}}{24} = - \frac{\ln{\left(\left|{u + 3}\right| \right)}}{24} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{24}$$

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$- \frac{\ln{\left(\left|{u + 3}\right| \right)}}{24} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{24} = - \frac{\ln{\left(\left|{u + 3}\right| \right)}}{24} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{24}$$

Recall that $$$v=u - 3$$$:

$$- \frac{\ln{\left(\left|{u + 3}\right| \right)}}{24} + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{24} = - \frac{\ln{\left(\left|{u + 3}\right| \right)}}{24} + \frac{\ln{\left(\left|{{\color{red}{\left(u - 3\right)}}}\right| \right)}}{24}$$

Recall that $$$u=4 e^{x}$$$:

$$\frac{\ln{\left(\left|{-3 + {\color{red}{u}}}\right| \right)}}{24} - \frac{\ln{\left(\left|{3 + {\color{red}{u}}}\right| \right)}}{24} = \frac{\ln{\left(\left|{-3 + {\color{red}{\left(4 e^{x}\right)}}}\right| \right)}}{24} - \frac{\ln{\left(\left|{3 + {\color{red}{\left(4 e^{x}\right)}}}\right| \right)}}{24}$$

Therefore,

$$\int{\frac{e^{- x}}{16 - 9 e^{- 2 x}} d x} = - \frac{\ln{\left(4 e^{x} + 3 \right)}}{24} + \frac{\ln{\left(\left|{4 e^{x} - 3}\right| \right)}}{24}$$

Add the constant of integration:

$$\int{\frac{e^{- x}}{16 - 9 e^{- 2 x}} d x} = - \frac{\ln{\left(4 e^{x} + 3 \right)}}{24} + \frac{\ln{\left(\left|{4 e^{x} - 3}\right| \right)}}{24}+C$$

$$$\int \frac{e^{- x}}{16 - 9 e^{- 2 x}}\, dx = \left(- \frac{\ln\left(4 e^{x} + 3\right)}{24} + \frac{\ln\left(\left|{4 e^{x} - 3}\right|\right)}{24}\right) + C$$$A