Integral of $$$\ln\left(\sqrt{x}\right)$$$

Find $$$\int \frac{\ln\left(x\right)}{2}\, dx$$$.

The input is rewritten: $$$\int{\ln{\left(\sqrt{x} \right)} d x}=\int{\frac{\ln{\left(x \right)}}{2} d x}$$$.

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 $$$f{\left(x \right)} = \ln{\left(x \right)}$$$:

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

For the integral $$$\int{\ln{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\ln{\left(x \right)}$$$ and $$$\operatorname{dv}=dx$$$.

Then $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d x}=x$$$ (steps can be seen »).

The integral becomes

$$\frac{{\color{red}{\int{\ln{\left(x \right)} d x}}}}{2}=\frac{{\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}}{2}=\frac{{\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}}{2}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

Therefore,

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

Simplify:

$$\int{\frac{\ln{\left(x \right)}}{2} d x} = \frac{x \left(\ln{\left(x \right)} - 1\right)}{2}$$

Add the constant of integration:

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

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