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?
The answe you got is wrong indeed, since

$$(x\ln x - x)' = 1\cdot \ln x + x\cdot \frac{1}{x} - 1 = \ln x + 1 - 1 = \ln x\neq x\ln x$$

To calculate the integral, use the per partes method. Since $$\int u(x)v'(x)dx = u(x)v(x) - \int u'(x) v(x)dx,$$ you should select $u$ such that you can calculate its derivative, while $v'$ should be such that you can calculate its antiderivative. In your case, it should be clear which of the functions $x$ and $\ln x$ is easier to find the antiderivative of.
You have to integrate the product $x\ln x$, the antiderivative of which is not immediate, and there is no obvious substitution to try.

So let us consider integration by parts. We can integrate the factor $x$ (but not $\ln x$), giving

$$\int x\ln x\,dx=\frac{x^2}2\ln x-\int\frac x2dx.$$

Bingo !