<s>[INST] 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 $f$.
 [/INST] 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: 
$$K\left(  \frac{1}{2}x^2\log x -\frac{1}{4}x^2 \right) + C$$
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