# How do you find the integral of 6x ln x dx ?

I found: $3 {x}^{2} \left[\ln \left(x\right) - \frac{1}{2}\right] + c$
$\int 6 x \ln \left(x\right) \mathrm{dx} = 6 {x}^{2} / 2 \ln \left(x\right) - \int 6 {x}^{\cancel{2}} / 2 \cdot \frac{1}{\cancel{x}} \mathrm{dx} =$
$= 3 {x}^{2} \ln \left(x\right) - \int 3 x \mathrm{dx} =$
$= 3 {x}^{2} \ln \left(x\right) - 3 {x}^{2} / 2 + c =$
$= 3 {x}^{2} \left[\ln \left(x\right) - \frac{1}{2}\right] + c$