Integral of $$$\frac{\ln\left(x\right)}{a^{4} x^{4}}$$$ with respect to $$$x$$$

Find $$$\int \frac{\ln\left(x\right)}{a^{4} x^{4}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{a^{4}}$$$ and $$$f{\left(x \right)} = \frac{\ln{\left(x \right)}}{x^{4}}$$$:

$${\color{red}{\int{\frac{\ln{\left(x \right)}}{a^{4} x^{4}} d x}}} = {\color{red}{\frac{\int{\frac{\ln{\left(x \right)}}{x^{4}} d x}}{a^{4}}}}$$

For the integral $$$\int{\frac{\ln{\left(x \right)}}{x^{4}} d x}$$$, 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 $$$\operatorname{dv}=\frac{dx}{x^{4}}$$$.

Then $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\frac{1}{x^{4}} d x}=- \frac{1}{3 x^{3}}$$$ (steps can be seen »).

Thus,

$$\frac{{\color{red}{\int{\frac{\ln{\left(x \right)}}{x^{4}} d x}}}}{a^{4}}=\frac{{\color{red}{\left(\ln{\left(x \right)} \cdot \left(- \frac{1}{3 x^{3}}\right)-\int{\left(- \frac{1}{3 x^{3}}\right) \cdot \frac{1}{x} d x}\right)}}}{a^{4}}=\frac{{\color{red}{\left(- \int{\left(- \frac{1}{3 x^{4}}\right)d x} - \frac{\ln{\left(x \right)}}{3 x^{3}}\right)}}}{a^{4}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{1}{3}$$$ and $$$f{\left(x \right)} = \frac{1}{x^{4}}$$$:

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:

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

Therefore,

$$\int{\frac{\ln{\left(x \right)}}{a^{4} x^{4}} d x} = \frac{- \frac{\ln{\left(x \right)}}{3 x^{3}} - \frac{1}{9 x^{3}}}{a^{4}}$$

Simplify:

$$\int{\frac{\ln{\left(x \right)}}{a^{4} x^{4}} d x} = \frac{- 3 \ln{\left(x \right)} - 1}{9 a^{4} x^{3}}$$

Add the constant of integration:

$$\int{\frac{\ln{\left(x \right)}}{a^{4} x^{4}} d x} = \frac{- 3 \ln{\left(x \right)} - 1}{9 a^{4} x^{3}}+C$$

$$$\int \frac{\ln\left(x\right)}{a^{4} x^{4}}\, dx = \frac{- 3 \ln\left(x\right) - 1}{9 a^{4} x^{3}} + C$$$A