Derivative of a distribution, including an integral of abs(x)

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I'm currently reading a book on finite element analysis and I lack knowledge on higher level of mathematics. My question is regarding the derivative of a distribution. I'm stuck on an example regarding the derivative of $|x|$. Since the book is written in french, I will summarize the example:

Basically, it is said that since $|x|$ is not differentiable at 0, we can associate a distribution thus:

$$<T'_{|x|},\phi> = \int_{-\infty}^{\infty} |x|\phi(x) dx$$

by definition, we can rewrite:

$$<T'_{|x|},\phi> = -<T_{|x|},\phi'>=\int_{-\infty}^{\infty} |x|\phi'(x)dx$$

Now, they seem to integrate by part but I'm not so sure which term they associated to $u,dv, v $ and $du$ in $\int udv = uv - \int vdu$. So I don't understand where the term $x\phi(x)|_{-\infty}^{0}$ and $x\phi(x)|_{0}^{\infty}$ are coming from:

$$=x\phi(x)|_{-\infty}^{0}-\int_{-\infty}^{0}\phi(x)dx -x\phi(x)|_{0}^{\infty}+\int_{0}^{\infty}\phi(x)dx$$

Afterwards, the the two terms I mentioned disappear and the rest of the example is:

$$=-\int_{-\infty}^{0}\phi(x)dx+\int_{0}^{\infty}\phi(x)dx=\int_{-\infty}^{\infty}g(x)\phi(x)dx$$ $$=<T_{g},\phi> \forall \phi\in\mathcal{D}(\Omega)$$

I believe I'm missing something in the integration by part. If you could help, I would greatly appreciate it.

Thank you very much.

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After a few trials, I finally understood why.

The reason I misused the integration by parts before is because I didn't use definite integral and then the term $uv$ was not definite either. By that I mean: $$\int_{a}^{b}udv=uv|_{a}^{b}-\int_{a}^{b}vdu$$

Also, $|x|$ was rewritten $-x\in]-\infty,0[$ and $x\in]0,\infty[$. Now it all makes sense.