Evaluate $\int_0^e{W(x)}\,\mathrm{d}x$

Integrate by parts with $u=W(x)$ and $\mathrm{d}v=\mathrm{d}x$

$$\int{W(x)}\,\mathrm{d}x=xW(x) - \int xW'(x)\,\mathrm{d}x=xW(x)-\int W(x)W'(x)e^{W(x)}\,\mathrm{d}x$$

Integrate by parts again with $u = W(x)$ and $\mathrm{d}v=W'(x)e^{W(x)}\,\mathrm{d}x$

$$\int W(x)W'(x)e^{W(x)}\,\mathrm{d}x=W(x)e^{W(x)} - \int W'(x)e^{W(x)}\,\mathrm{d}x=W(x)e^{W(x)}-e^{W(x)}$$

So $$\int{W(x)}\,\mathrm{d}x=xW(x) -W(x)e^{W(x)}+e^{W(x)}=x\left(W(x)-1+\frac{1}{W(x)}\right)$$

Note that $W(e) = 1$ and $W(0)=0$ and $\displaystyle{\lim_{x\to 0}\frac{x}{W(x)}=1}$

$$\left|x\left(W(x)-1+\frac{1}{W(x)}\right)\right|_0^e=e\left(W(e)-1+\frac{1}{W(e)}\right) - \left(\lim_{x\to 0}\frac{x}{W(x)}\right) = e-1$$