Integral of $$$\frac{\ln\left(x\right)}{x^{3}}$$$

Find $$$\int \frac{\ln\left(x\right)}{x^{3}}\, dx$$$.

For the integral $$$\int{\frac{\ln{\left(x \right)}}{x^{3}} d x}$$$, 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 $$$\operatorname{dv}=\frac{dx}{x^{3}}$$$.

Then $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\frac{1}{x^{3}} d x}=- \frac{1}{2 x^{2}}$$$ (steps can be seen »).

The integral can be rewritten as

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

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 $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$:

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

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

Therefore,

$$\int{\frac{\ln{\left(x \right)}}{x^{3}} d x} = - \frac{\ln{\left(x \right)}}{2 x^{2}} - \frac{1}{4 x^{2}}$$

Simplify:

$$\int{\frac{\ln{\left(x \right)}}{x^{3}} d x} = \frac{- 2 \ln{\left(x \right)} - 1}{4 x^{2}}$$

Add the constant of integration:

$$\int{\frac{\ln{\left(x \right)}}{x^{3}} d x} = \frac{- 2 \ln{\left(x \right)} - 1}{4 x^{2}}+C$$

$$$\int \frac{\ln\left(x\right)}{x^{3}}\, dx = \frac{- 2 \ln\left(x\right) - 1}{4 x^{2}} + C$$$A