$$\int_{-\infty}^{\infty} \psi^{\dagger}\psi dx = \int_{-\infty}^{\infty} (A e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(A e^{-\lambda |x|} e^{-i\omega t})dx$$
$$= A^{\dagger}A\int_{-\infty}^{\infty} (e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(e^{-\lambda |x|} e^{-i\omega t})dx$$
Where $\dagger$ represents the Hermitian conjugate, or the complex conjugate in the case of $A$, so $$A^{\dagger}A = |A|^2$$
and that is where the $|A|^2$ comes from, regardless of whether $A$ is real or not.
Perhaps its a little clearer if you shorten the contents of the brackets (and lets drop the constants too):
$$\frac{d\langle x\rangle }{dt} \propto \int _{-\infty} ^{\infty} x \frac{\partial }{\partial x} \left[ \ldots\right] dx$$
$$ = \int _{-\infty} ^{\infty} \frac{\partial }{\partial x} \left(x \left[ \ldots\right] \right)dx - \int _{-\infty} ^{\infty} \frac{\partial x}{\partial x} \left[ \ldots\right] dx$$
The first term is the boundary term, which as you discussed with ACuriousMind, goes to zero. More importantly, the derivative on $[\ldots]$ has shifted to $x$ on the second term
Where of course $\frac{\partial x}{\partial x} = 1$
And hence the result:
$$\frac{d\langle x\rangle }{dt} = -\frac{i\hbar}{2m} \int _{-\infty} ^{\infty} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^* }{\partial x}\Psi \right) dx$$
Edit:
To the last question: No, that statement is not true in general. Only sometimes are we allowed to throw away the boundary term (it depends on the physical situation, not the maths).
Do you remember how integration by parts comes about? From the product rule:
$$
\frac{d(fg)}{dx} = \frac{df}{dx}g + f\frac{dg}{dx}
$$
Then rearrange:
$$\frac{df}{dx}g = \frac{dfg}{dx} - f\frac{dg}{dx}$$
Then integrate both sides:
$$\int_{x_0}^{x_1}\frac{df}{dx} g dx = \int_{x_0}^{x_1}\frac{d(fg)}{dx}dx - \int_{x_0}^{x_1} f\frac{dg}{dx}dx$$
Where the derivative and the integral on the first term of the RHS cancel to become
$$\int_{x_0}^{x_1}\frac{df}{dx} g dx = fg |_{x_0}^{x_1}- \int_{x_0}^{x_1} f\frac{dg}{dx}dx$$
What we're saying is that we know $fg$ at $x_0$ and $x_1$ (ie the boundaries) is zero, so we drop that term. Then the net result is we've switched the derivative from f to g, at the expense of a sign change.
Best Answer
You should try solving this on a piece of paper. I don't think that you understand correctly how the author passes from $\partial/\partial t$ to $\partial^2/\partial^2 x$. You can not use Schrodinger's equation for $|\Psi|^2$, it is only valid for $\Psi(x, t)$:
$$ i \hbar \frac{\partial}{\partial t} \Psi(x, t) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} \Psi(x, t). $$
For $\Psi^*$ another equation holds, which can be obtained by conjugating both sides of the Schrodinger's equation:
$$ - i \hbar \frac{\partial}{\partial t} \Psi^*(x, t) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} \Psi^*(x, t). $$
Note the minus sign on the l.h.s. It comes from conjugating an imaginary unit ($i^* = -i$). This minus sign is responsible for the minus sign in your answer.
Now I trust you to carefully expand both parts of your equation and see for yourself that they are indeed equal.