Find the following integral: $\int {{{(\ln x)}^2}} dx$ by using the method of integration by parts 
Find $\int {{{(\ln x)}^2}} dx$ by using the method of integration by parts.

My attempt:
$$\eqalign{
  & \int {{{(\ln x)}^2}} dx = \int {2\ln x} dx  \cr 
  & u = \ln x,{\rm{ }}{{du} \over {dx}} = {1 \over x}  \cr 
  & {{dv} \over {dx}} = 2,{\rm{ }}v = 2x  \cr 
  & so:  \cr 
  & \int {{{(\ln x)}^2}} dx = 2x\ln x - \int {2x \times {1 \over x}} dx  \cr 
  &  = 2x\ln x - 2x + C \cr} $$

This is the wrong answer, the right answer is:
$$x{(\ln x)^2} - 2x\ln x + 2x + C$$
What have I missed?

Hint: try $u = \ln x$ and $\frac{dv}{dx} = \ln x$. Then solve $dv = \ln x dx$ by integration by parts exactly as you have it above. 
