Question about a integral step(integration method) Let X have the pdf 
 $f(n) =
\begin{cases}
x e ^{-x},  & \text{$0 \leq x < \infty$} \\
0, & \text{elsewhere}
\end{cases}$
then
$\begin{align} M(t) &=\int^{b}_{0} x e^{-(1-t)x} dx  \\ 
& =\lim_{b \rightarrow \infty} [-\frac{x e^{-(1-t)x}}{1-t}-\frac{e^{-(1-t)x}}{(1-t)^2}]^{b}_{0} \\ 
& = \frac{1}{(1-t)^2} \end{align}$ 

Question: For the integration, what kind of integration method is that? is it integration by parts? 

 A: Yes, indeed. Integration by parts can be written as.
$$\int [f\cdot g'](x)\mathrm d x = [f\cdot g](x)-\int [f'\cdot g](x)\mathrm d x$$
So taking $f(x):=x$ and $g(x):=s^{-1}\mathrm e^{sx}$, so that $f'(x)=1$ and $g'(x)=\mathrm e^{sx}$, 
$$\begin{align}\int x\mathrm  e^{sx}~\mathrm d x &=xs^{-1}\mathrm e^{sx}-s^{-1}\int \mathrm e^{sx}~\mathrm d x\\&=xs^{-1}\mathrm e^{sx}-s^{-2}\mathrm e^{sx}+C\end{align}$$
And so when $s:=-(1-t)$ we obtain the indefinite integration.
$$\begin{align}\int x\mathrm e^{-(1-t)~x}~\mathrm d x &= -\dfrac{x~\mathrm e^{-(1-t)~x}}{(1-t)}-\dfrac{\mathrm e^{-(1-t)~x}}{(1-t)^2}+C\\&=-\dfrac{((1-t)x+1)~\mathrm e^{-(1-t)~x}}{(1-t)^2}+C\end{align}$$
And the definite integration over the domain $[0..b)$ becomes
$$\begin{align}\int_0^b x\mathrm e^{-(1-t)~x}~\mathrm d x &= \left[-\dfrac{x~\mathrm e^{-(1-t)~x}}{(1-t)}-\dfrac{\mathrm e^{-(1-t)~x}}{(1-t)^2}\right]_{x=0}^{x=b}\\&=-\dfrac{((1-t)b+1)~\mathrm e^{-(1-t)~b}}{(1-t)^2}+\dfrac{1}{(1-t)^2}\end{align}$$
And so, because the exponential tends down to zero faster than the coefficient tends towards infinitude, taking it to the limit $b\to\infty$ gets us:
$$\begin{align}\int_0^\infty x~\mathrm e^{-(1-t)~x}~\mathrm d x=\dfrac{1}{(1-t)^2}\end{align}$$
