Integral of $$$2 u \ln\left(2 u\right) - u - 2$$$

Find $$$\int \left(2 u \ln\left(2 u\right) - u - 2\right)\, du$$$.

Integrate term by term:

$${\color{red}{\int{\left(2 u \ln{\left(2 u \right)} - u - 2\right)d u}}} = {\color{red}{\left(- \int{2 d u} - \int{u d u} + \int{2 u \ln{\left(2 u \right)} d u}\right)}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=2$$$:

$$- \int{u d u} + \int{2 u \ln{\left(2 u \right)} d u} - {\color{red}{\int{2 d u}}} = - \int{u d u} + \int{2 u \ln{\left(2 u \right)} d u} - {\color{red}{\left(2 u\right)}}$$

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

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = u \ln{\left(2 u \right)}$$$:

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

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

Let $$$\operatorname{\omega}=\ln{\left(2 u \right)}$$$ and $$$\operatorname{dv}=u du$$$.

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

Therefore,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = u$$$:

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

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

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

Therefore,

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

Simplify:

$$\int{\left(2 u \ln{\left(2 u \right)} - u - 2\right)d u} = u \left(u \left(\ln{\left(u \right)} + \ln{\left(2 \right)}\right) - u - 2\right)$$

Add the constant of integration:

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

$$$\int \left(2 u \ln\left(2 u\right) - u - 2\right)\, du = u \left(u \left(\ln\left(u\right) + \ln\left(2\right)\right) - u - 2\right) + C$$$A