Integral of $$$- \sqrt{x} + \frac{1}{x}$$$

Find $$$\int \left(- \sqrt{x} + \frac{1}{x}\right)\, dx$$$.

Integrate term by term:

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

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

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{1}{2}$$$:

$$\ln{\left(\left|{x}\right| \right)} - {\color{red}{\int{\sqrt{x} d x}}}=\ln{\left(\left|{x}\right| \right)} - {\color{red}{\int{x^{\frac{1}{2}} d x}}}=\ln{\left(\left|{x}\right| \right)} - {\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=\ln{\left(\left|{x}\right| \right)} - {\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}$$

Therefore,

$$\int{\left(- \sqrt{x} + \frac{1}{x}\right)d x} = - \frac{2 x^{\frac{3}{2}}}{3} + \ln{\left(\left|{x}\right| \right)}$$

Add the constant of integration:

$$\int{\left(- \sqrt{x} + \frac{1}{x}\right)d x} = - \frac{2 x^{\frac{3}{2}}}{3} + \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \left(- \sqrt{x} + \frac{1}{x}\right)\, dx = \left(- \frac{2 x^{\frac{3}{2}}}{3} + \ln\left(\left|{x}\right|\right)\right) + C$$$A