A more precise notation is this one
$$\int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx=\left(
u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right) -\int_{x_{1}}^{x_{2}}u^{\prime
}(x)v(x)dx$$
which is derived from the derivative rule for the product
$$(u(x)v(x))^{\prime }=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$
or
$$u(x)v^{\prime }(x)=(u(x)v(x))^{\prime }-u^{\prime }(x)v(x).$$
So
$$\begin{eqnarray*}
\int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx
&=&\int_{x_{1}}^{x_{2}}(u(x)v(x))^{\prime }dx-\int_{x_{1}}^{x_{2}}u^{\prime
}(x)v(x)dx \\
&=&\left. (u(x)v(x))\right\vert
_{x=x_{1}}^{x=x_{2}}-\int_{x_{1}}^{x_{2}}u(x)v(x)dx \\
&=&\left( u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right)
-\int_{x_{1}}^{x_{2}}u^{\prime }(x)v(x)dx.
\end{eqnarray*}.$$
If you write $dv=v^{\prime }(x)dx$ and $du=u^{\prime }(x)dx$, you get your
formula but with $u,v$ as a function of $x$
$$\int_{v_{1}(x)}^{v_{2}(x)}u(x)dv=\left(
u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right)
-\int_{u_{1}(x)}^{u_{2}(x)}v(x)du$$
Example: Assume you want to evaluate $\int_{x_{1}}^{x_{2}}\log
xdx=\int_{x_{1}}^{x_{2}}1\cdot \log xdx$. You can choose $v^{\prime }(x)=1$
and $u(x)=\log x$. Then $v(x)=x$ (omitting the constant of integration) and
$u^{\prime }(x)=\frac{1}{x}$. Hence
$$\begin{eqnarray*}
\int_{x_{1}}^{x_{2}}\log xdx &=&\int_{x_{1}}^{x_{2}}1\cdot \log xdx \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)
-\int_{x_{1}}^{x_{2}}\frac{1}{x}\cdot xdx \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)
-\int_{x_{1}}^{x_{2}}dx \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left(
x_{2}-x_{1}\right)
\end{eqnarray*}$$
The same example with your formula:
$$u=\log x,v=x,dv=dx,v=x,du=\frac{1}{x}dx$$
$$u_{2}=\log x_{2},u_{1}=\log x_{1},v_{2}=x_{2},v_{1}=x_{1}$$
$$\begin{eqnarray*}
\int_{v_{1}}^{v_{2}}udv &=&\left( u_{2}v_{2}-u_{1}v_{2}\right)
-\int_{u_{1}}^{u_{2}}vdu \\
\int_{x_{1}}^{x_{2}}\log xdx &=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot
x_{1}\right) -\int_{\log x_{1}}^{\log x_{2}}xdu \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)
-\int_{x_{1}}^{x_{2}}x\cdot \frac{1}{x}dx \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left(
x_{2}-x_{1}\right).
\end{eqnarray*}$$
Note: The limits of integration, although different in terms of $u(x),v(x)$, when expressed in terms of the same variable $x$ of functions $u(x),v(x)$ are the same in both sides.
For a strategy on how to chose the $u$ and $v$ terms see this question.
Let $f(x)=x^{s-1}e^{-x}, F(y)= \int_0^y f(x)dx$, $G(y) = F(ny)$ then $G'(x) =n F'(nx)= n f(nx)$ thus $$\int_0^y n f(nx)dx = G(y)-G(0) = F(ny)=\int_0^{ny}f(x)dx$$
Letting $y\to \infty$ you get the change of variable formula $$\Gamma(s)=\int_0^\infty f(x)dx=n\int_0^\infty f(nx)dx= n \int_0^\infty n^{s-1}x^{s-1}e^{-nx}dx$$
Best Answer
What you are looking for is $$\frac {du}{dt}=\frac d{dt}\left(\left(1-\frac tn\right)^n\right)=n\left(1-\frac tn\right)^{n-1}\times\frac d{dt}\left(1-\frac tn\right)$$by the chain rule$$=n\left(1-\frac tn\right)^{n-1}\times-\frac1n=-\left(1-\frac tn\right)^{n-1}$$which should be as required for your integration. (I notice they have written this term with a $\frac nn$ in front, which is basically redundant.)