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I have the following differential equation

$$\frac{df}{dx} = Kx\log x,$$

$K$ a constant. I'm wandering how one might solve for $f$.

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    $\begingroup$ Integrate both sides with respect to $x$. $\endgroup$ – Git Gud May 19 '14 at 10:40
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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!

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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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$$ \int df = \int k x \ln x ~dx \implies f =k \left[ (\ln x)(x^2)/2 - \int x/2 ~dx \right] \implies f = k \left[ (\ln x)(x^2)/2 -(x^2)/4\right] + I$$

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