Integral of $$$\frac{\sqrt{2 - x}}{2 x}$$$

Find $$$\int \frac{\sqrt{2 - x}}{2 x}\, dx$$$.

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

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

Let $$$u=\sqrt{2 - x}$$$.

Then $$$du=\left(\sqrt{2 - x}\right)^{\prime }dx = - \frac{1}{2 \sqrt{2 - x}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{\sqrt{2 - x}} = - 2 du$$$.

The integral becomes

$$\frac{{\color{red}{\int{\frac{\sqrt{2 - x}}{x} d x}}}}{2} = \frac{{\color{red}{\int{\left(- \frac{2 u^{2}}{2 - u^{2}}\right)d u}}}}{2}$$

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

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

Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division (steps can be seen »):

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$- \int{\frac{2}{2 - u^{2}} d u} + {\color{red}{\int{1 d u}}} = - \int{\frac{2}{2 - u^{2}} d u} + {\color{red}{u}}$$

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

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

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

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

Integrate term by term:

$$u - 2 {\color{red}{\int{\left(\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} - \frac{\sqrt{2}}{4 \left(u - \sqrt{2}\right)}\right)d u}}} = u - 2 {\color{red}{\left(- \int{\frac{\sqrt{2}}{4 \left(u - \sqrt{2}\right)} d u} + \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u}\right)}}$$

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

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

Let $$$v=u - \sqrt{2}$$$.

Then $$$dv=\left(u - \sqrt{2}\right)^{\prime }du = 1 du$$$ (steps can be seen »), and we have that $$$du = dv$$$.

So,

$$u - 2 \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u} + \frac{\sqrt{2} {\color{red}{\int{\frac{1}{u - \sqrt{2}} d u}}}}{2} = u - 2 \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u} + \frac{\sqrt{2} {\color{red}{\int{\frac{1}{v} d v}}}}{2}$$

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

$$u - 2 \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u} + \frac{\sqrt{2} {\color{red}{\int{\frac{1}{v} d v}}}}{2} = u - 2 \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u} + \frac{\sqrt{2} {\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

Recall that $$$v=u - \sqrt{2}$$$:

$$u + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} - 2 \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u} = u + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(u - \sqrt{2}\right)}}}\right| \right)}}{2} - 2 \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u}$$

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

$$u + \frac{\sqrt{2} \ln{\left(\left|{u - \sqrt{2}}\right| \right)}}{2} - 2 {\color{red}{\int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u}}} = u + \frac{\sqrt{2} \ln{\left(\left|{u - \sqrt{2}}\right| \right)}}{2} - 2 {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{u + \sqrt{2}} d u}}{4}\right)}}$$

Let $$$v=u + \sqrt{2}$$$.

Then $$$dv=\left(u + \sqrt{2}\right)^{\prime }du = 1 du$$$ (steps can be seen »), and we have that $$$du = dv$$$.

So,

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

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

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

Recall that $$$v=u + \sqrt{2}$$$:

$$u + \frac{\sqrt{2} \ln{\left(\left|{u - \sqrt{2}}\right| \right)}}{2} - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = u + \frac{\sqrt{2} \ln{\left(\left|{u - \sqrt{2}}\right| \right)}}{2} - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(u + \sqrt{2}\right)}}}\right| \right)}}{2}$$

Recall that $$$u=\sqrt{2 - x}$$$:

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

Therefore,

$$\int{\frac{\sqrt{2 - x}}{2 x} d x} = \sqrt{2 - x} + \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2 - x} - \sqrt{2}}\right| \right)}}{2} - \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2 - x} + \sqrt{2}}\right| \right)}}{2}$$

Add the constant of integration:

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

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