Integral of $$$\ln\left(2 x^{3}\right)$$$

Find $$$\int \ln\left(2 x^{3}\right)\, dx$$$.

For the integral $$$\int{\ln{\left(2 x^{3} \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\ln{\left(2 x^{3} \right)}$$$ and $$$\operatorname{dv}=dx$$$.

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

The integral becomes

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=3$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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