Integral of $$$8 x \ln\left(4 x^{2}\right)$$$

Find $$$\int 8 x \ln\left(4 x^{2}\right)\, dx$$$.

Let $$$u=4 x^{2}$$$.

Then $$$du=\left(4 x^{2}\right)^{\prime }dx = 8 x dx$$$ (steps can be seen »), and we have that $$$x dx = \frac{du}{8}$$$.

The integral can be rewritten as

$${\color{red}{\int{8 x \ln{\left(4 x^{2} \right)} d x}}} = {\color{red}{\int{\ln{\left(u \right)} d u}}}$$

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

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

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

Thus,

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

Recall that $$$u=4 x^{2}$$$:

$$- {\color{red}{u}} + {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} = - {\color{red}{\left(4 x^{2}\right)}} + {\color{red}{\left(4 x^{2}\right)}} \ln{\left({\color{red}{\left(4 x^{2}\right)}} \right)}$$

Therefore,

$$\int{8 x \ln{\left(4 x^{2} \right)} d x} = 4 x^{2} \ln{\left(4 x^{2} \right)} - 4 x^{2}$$

Simplify:

$$\int{8 x \ln{\left(4 x^{2} \right)} d x} = 4 x^{2} \left(2 \ln{\left(x \right)} - 1 + 2 \ln{\left(2 \right)}\right)$$

Add the constant of integration:

$$\int{8 x \ln{\left(4 x^{2} \right)} d x} = 4 x^{2} \left(2 \ln{\left(x \right)} - 1 + 2 \ln{\left(2 \right)}\right)+C$$

$$$\int 8 x \ln\left(4 x^{2}\right)\, dx = 4 x^{2} \left(2 \ln\left(x\right) - 1 + 2 \ln\left(2\right)\right) + C$$$A