Integral of $$$x^{4} \ln\left(x\right)$$$

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

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

Then $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{x^{4} d x}=\frac{x^{5}}{5}$$$ (steps can be seen »).

So,

$${\color{red}{\int{x^{4} \ln{\left(x \right)} d x}}}={\color{red}{\left(\ln{\left(x \right)} \cdot \frac{x^{5}}{5}-\int{\frac{x^{5}}{5} \cdot \frac{1}{x} d x}\right)}}={\color{red}{\left(\frac{x^{5} \ln{\left(x \right)}}{5} - \int{\frac{x^{4}}{5} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{5}$$$ and $$$f{\left(x \right)} = x^{4}$$$:

$$\frac{x^{5} \ln{\left(x \right)}}{5} - {\color{red}{\int{\frac{x^{4}}{5} d x}}} = \frac{x^{5} \ln{\left(x \right)}}{5} - {\color{red}{\left(\frac{\int{x^{4} d x}}{5}\right)}}$$

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

$$\frac{x^{5} \ln{\left(x \right)}}{5} - \frac{{\color{red}{\int{x^{4} d x}}}}{5}=\frac{x^{5} \ln{\left(x \right)}}{5} - \frac{{\color{red}{\frac{x^{1 + 4}}{1 + 4}}}}{5}=\frac{x^{5} \ln{\left(x \right)}}{5} - \frac{{\color{red}{\left(\frac{x^{5}}{5}\right)}}}{5}$$

Therefore,

$$\int{x^{4} \ln{\left(x \right)} d x} = \frac{x^{5} \ln{\left(x \right)}}{5} - \frac{x^{5}}{25}$$

Simplify:

$$\int{x^{4} \ln{\left(x \right)} d x} = \frac{x^{5} \left(5 \ln{\left(x \right)} - 1\right)}{25}$$

Add the constant of integration:

$$\int{x^{4} \ln{\left(x \right)} d x} = \frac{x^{5} \left(5 \ln{\left(x \right)} - 1\right)}{25}+C$$

$$$\int x^{4} \ln\left(x\right)\, dx = \frac{x^{5} \left(5 \ln\left(x\right) - 1\right)}{25} + C$$$A