How do you integrate int ln x^2 dx  using integration by parts?

$\int \ln {x}^{2} \mathrm{dx} = x \cdot \ln {x}^{2} - 2 x + C$

Explanation:

The formula $\int u \cdot \mathrm{dv} = u \cdot v - \int v \cdot \mathrm{du}$

the given $\int \ln {x}^{2} \mathrm{dx}$
Let $u = \ln {x}^{2}$
$\mathrm{dv} = \mathrm{dx}$

$v = x$

$\mathrm{du} = 2 x \cdot \frac{\mathrm{dx}}{x} ^ 2 = 2 \cdot \frac{\mathrm{dx}}{x}$

$\int \ln {x}^{2} \mathrm{dx} = x \cdot \ln {x}^{2} - \int x \cdot 2 \cdot \frac{\mathrm{dx}}{x} + C$

$\int \ln {x}^{2} \mathrm{dx} = x \cdot \ln {x}^{2} - \int 2 \cdot \mathrm{dx} + C$

$\int \ln {x}^{2} \mathrm{dx} = x \cdot \ln {x}^{2} - 2 x + C$