Integral of $$$\frac{\sqrt{2}}{2 \left(- x^{2} - 6 x + 7\right)}$$$

Find $$$\int \frac{\sqrt{2}}{2 \left(- x^{2} - 6 x + 7\right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{\sqrt{2}}{2}$$$ and $$$f{\left(x \right)} = \frac{1}{- x^{2} - 6 x + 7}$$$:

$${\color{red}{\int{\frac{\sqrt{2}}{2 \left(- x^{2} - 6 x + 7\right)} d x}}} = {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{- x^{2} - 6 x + 7} d x}}{2}\right)}}$$

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

$$\frac{\sqrt{2} {\color{red}{\int{\frac{1}{- x^{2} - 6 x + 7} d x}}}}{2} = \frac{\sqrt{2} {\color{red}{\int{\left(\frac{1}{8 \left(x + 7\right)} - \frac{1}{8 \left(x - 1\right)}\right)d x}}}}{2}$$

Integrate term by term:

$$\frac{\sqrt{2} {\color{red}{\int{\left(\frac{1}{8 \left(x + 7\right)} - \frac{1}{8 \left(x - 1\right)}\right)d x}}}}{2} = \frac{\sqrt{2} {\color{red}{\left(- \int{\frac{1}{8 \left(x - 1\right)} d x} + \int{\frac{1}{8 \left(x + 7\right)} d x}\right)}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 1}$$$:

$$\frac{\sqrt{2} \left(\int{\frac{1}{8 \left(x + 7\right)} d x} - {\color{red}{\int{\frac{1}{8 \left(x - 1\right)} d x}}}\right)}{2} = \frac{\sqrt{2} \left(\int{\frac{1}{8 \left(x + 7\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x - 1} d x}}{8}\right)}}\right)}{2}$$

Let $$$u=x - 1$$$.

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

Therefore,

$$\frac{\sqrt{2} \left(\int{\frac{1}{8 \left(x + 7\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x - 1} d x}}}}{8}\right)}{2} = \frac{\sqrt{2} \left(\int{\frac{1}{8 \left(x + 7\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8}\right)}{2}$$

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

$$\frac{\sqrt{2} \left(\int{\frac{1}{8 \left(x + 7\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8}\right)}{2} = \frac{\sqrt{2} \left(\int{\frac{1}{8 \left(x + 7\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}\right)}{2}$$

Recall that $$$u=x - 1$$$:

$$\frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \int{\frac{1}{8 \left(x + 7\right)} d x}\right)}{2} = \frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{8} + \int{\frac{1}{8 \left(x + 7\right)} d x}\right)}{2}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(x \right)} = \frac{1}{x + 7}$$$:

$$\frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x - 1}\right| \right)}}{8} + {\color{red}{\int{\frac{1}{8 \left(x + 7\right)} d x}}}\right)}{2} = \frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x - 1}\right| \right)}}{8} + {\color{red}{\left(\frac{\int{\frac{1}{x + 7} d x}}{8}\right)}}\right)}{2}$$

Let $$$u=x + 7$$$.

Then $$$du=\left(x + 7\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

Thus,

$$\frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x - 1}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{x + 7} d x}}}}{8}\right)}{2} = \frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x - 1}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8}\right)}{2}$$

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

$$\frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x - 1}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8}\right)}{2} = \frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x - 1}\right| \right)}}{8} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}\right)}{2}$$

Recall that $$$u=x + 7$$$:

$$\frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x - 1}\right| \right)}}{8} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8}\right)}{2} = \frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x - 1}\right| \right)}}{8} + \frac{\ln{\left(\left|{{\color{red}{\left(x + 7\right)}}}\right| \right)}}{8}\right)}{2}$$

Therefore,

$$\int{\frac{\sqrt{2}}{2 \left(- x^{2} - 6 x + 7\right)} d x} = \frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x - 1}\right| \right)}}{8} + \frac{\ln{\left(\left|{x + 7}\right| \right)}}{8}\right)}{2}$$

Simplify:

$$\int{\frac{\sqrt{2}}{2 \left(- x^{2} - 6 x + 7\right)} d x} = \frac{\sqrt{2} \left(- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 7}\right| \right)}\right)}{16}$$

Add the constant of integration:

$$\int{\frac{\sqrt{2}}{2 \left(- x^{2} - 6 x + 7\right)} d x} = \frac{\sqrt{2} \left(- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 7}\right| \right)}\right)}{16}+C$$

$$$\int \frac{\sqrt{2}}{2 \left(- x^{2} - 6 x + 7\right)}\, dx = \frac{\sqrt{2} \left(- \ln\left(\left|{x - 1}\right|\right) + \ln\left(\left|{x + 7}\right|\right)\right)}{16} + C$$$A