Integration by parts of $x\ln (x)$ is giving me the wrong answer? $\newcommand{\d}[1]{\mathop{\mathrm d #1}}$
To integrate $x\ln (x)$, let $u = x$ and $\d v = \ln(x)\d x$. It becomes
$$  \int x\ln(x)\d x = x\int \ln(x)\d x\,-\int \int \ln(x) \d {x^2} $$
simplified to
$$ \int x\ln(x)\d x = x(x\ln(x) -x)\,-\int(x\ln(x) - x)\d x $$
break up integral and solve integral $x \d x$
$$ \int x\ln(x)\d x = x(x\ln(x) -x)\,-\int x\ln(x) \d x - \frac{1}{2}x^2 \qquad(\star)$$
add (integral $x \ln x \d x$) to both sides then divide by two
$$ \int x\ln(x)dx = \frac{x(x\ln(x) -x) - \frac{1}{2}x^2}{2}$$
and this is apparently not the answer. The answer is actually 
$$\frac{1}{4}x^2(-1+2\log (x))$$
which is not the same function.
What am I doing wrong?
I can do this problem fine if I assign $u$ and $v$ the other way around, but this should still work and I feel like I'm making a dumb mistake somewhere.

$$\begin{cases} du=xdx \\ v=\ln { x }  \end{cases}\Rightarrow \begin{cases} u=\frac { { x }^{ 2 } }{ 2 }  \\ dv=\frac { dx }{ x }  \end{cases}\\ \int { x\ln { xdx }  } =\frac { { x }^{ 2 }\ln { x }  }{ 2 } -\int { \frac { x }{ 2 } dx= } \frac { { x }^{ 2 }\ln { x }  }{ 2 } -\frac { { x }^{ 2 } }{ 4 } +C$$
