Solving $df/dx = x\log x$ I have the following differential equation
$$\frac{df}{dx} = Kx\log x,$$
$K$ a constant. I'm wandering how one might solve for $f$.
 A: Hint (or complete answer, maybe):  your equation is separable, i.e., it can be written in the form $f'(x) = K \, g(x)$, where $'$ denotes differentiation with respect to $x$. Therefore, integrating both sides, $f(x) = K \, G(x) + C$, where $G$ is a primitive of $g$.
The only tricky thing here is integrating the function $x \log{x}$ which is given by:
$$G(x) = \int x \log{x} \, \mathrm{d}x = \frac{x^2}{2} \log{x} - \int \frac{x^2}{2} \frac{1}{x} \mathrm{d}x.$$
I'm sure you can take it from here.
Cheers!
A: Integrating by parts we know that: 
$$fg = \int f'g+\int fg'$$
Now take $f' = x,\; g=\log x$ (I use $e$ as the base for the logarithm). This means that $f= \frac{1}{2}x^2$ and $g'= \frac{1}{x}$ , so we get:
$$\int f'g = \int x \log x \;dx = \frac{1}{2}x^2\log x - \int \frac{1}{2}x^2\frac{1}{x}\;dx$$
$$= \frac{1}{2}x^2\log x - \frac{1}{2}\int x\;dx = \frac{1}{2}x^2\log x -\frac{1}{4}x^2 + C$$
So the answer is: 
$$K\left(  \frac{1}{2}x^2\log x -\frac{1}{4}x^2 \right) + C$$
A: $$ \int df = \int k x \ln x ~dx
 \implies f =k \left[ (\ln x)(x^2)/2  - \int x/2 ~dx \right]
\implies f = k \left[ (\ln x)(x^2)/2 -(x^2)/4\right] + I$$
