Can anyone help showing why my calculation of $\int x\ln x$ $dx$ is wrong? 
Suppose $\frac{dy}{dx}=x\ln x,$ my teacher asks me to find $y$.

So I assume I got to integrate the right hand side:
$$\int x\ln x\, dx$$
The result I got is 
$$
\int x\ln x\, dx=x\ln x-x+C\tag{1}
$$ 
But, apparently, it is wrong since taking the derivative gives:
$$
(x\ln x-x+C)'=\ln x+1-1=\ln x.
$$
Can you please give me a hand?

Hint:
Use integration by parts and be careful, you integrated $\ln(x)$ but you have to integrate $\color{red}{x} \cdot \ln(x)$.
