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?
 A: By integration by parts,
$$\int (\ln(x))^2 dx = x (\ln(x))^2 - \int x d \left((\ln(x))^2\right)$$
$$\int x d \left((\ln(x))^2\right) = \int x \cdot 2 \ln(x) \dfrac{dx}x = 2\int \ln(x)dx$$
Now again by integration by parts, we have
$$\int \ln(x)dx = \ln(x) \cdot x - \int x \cdot \dfrac{dx}x = x \ln(x) - x$$
Putting all this together, we get
$$\int (\ln(x))^2 dx = x (\ln(x))^2-2x\ln(x) + 2x + \text{constant}$$
A: 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. 
A: $$
\begin{align}
\int\log(x)^2\,\mathrm{d}x
&=x\log(x)^2-\int 2\log(x)\,\mathrm{d}x\tag{$x$ and $\log(x)^2$}\\
&=x\log(x)^2-2x\log(x)+\int2\,\mathrm{d}x\tag{$x$ and $\log(x)$}\\
&=x\log(x)^2-2x\log(x)+2x+C
\end{align}
$$
A: Hint: Apply partial integration (viz $\displaystyle u(x) v(x) = \int u(x) v'(x)\, \mathrm dx + \int u'(x) v(x) \,\mathrm dx$) to $u = \log x, v' = \log x$.
A: $$\int {{{(\log x)}^2}} dx $$
(To remind that $(\log x)^2\ne \log x^2; (\log x)^2=(\log x)\cdot(\log x)$ while $\log x^2=2\log x$)
$$\int {{{(\log x)}.(\log x)}} dx $$
$$(\log x)\cdot\int {{(\log x)}} dx -\int [{d/dx(\log x)}{\int {\log x}dx} ]dx$$
since $$\int {\log x}dx=x\log x-x$$
so
$$(\log x)\cdot(x\log x - x) -\int [\frac {1}{x}(x\log x - x)]dx$$
$$(\log x)\cdot(x\log x - x) -\int {(\log x - 1)} dx$$
$$x(\log x)^2 - x\log x -(x\log x - x)+x$$
$$x(\log x)^2 - 2x\log x +2x+C$$
