Integral of $$$\ln\left(\frac{a^{2}}{x^{2}}\right)$$$ with respect to $$$x$$$

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

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

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

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

The integral becomes

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=-2$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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