How does Wolfram get from the first form to the second alternate form? So, I was trying to compute an integral but I couldn't actually manage getting anywhere with it in its initial form. So, I inserted the function in Wolfram Alpha and I really got a nicer form (second alternate form). But I want to understand how that was done, I really need some help here.

Input:
  $\dfrac{1}{(e^t + 1)(e^t + 2)(e^t + 3)}$
Alternate forms:
  $\dfrac{1}{11e^t + 6e^{2t} + e^{3t} + 6}\\
-\dfrac{1}{e^t + 2} + \dfrac{1}{2(e^t + 3)} + \dfrac{1}{2(e^t + 1)}
$


One trick to help deriving the $2^{nd}$ alternate form is following identity:
$$\frac{1}{(z+a)(z+b)} = \frac{1}{b-a}\left(\frac{1}{z+a} - \frac{1}{z+b}\right)$$
In particular, when $a,b$ differs by an integer, it reduces to a pattern relatively easy
to detect and apply this trick. e.g.
$$\begin{align}
\frac{1}{(z+1)(z+2)(z+3)} 
&= \left(\frac{1}{z+1}-\frac{1}{z+2}\right)\frac{1}{z+3}\\
&= \frac{1}{(z+1)(z+3)} - \frac{1}{(z+2)(z+3)}\\
&= \frac12\left(\frac{1}{z+1}-\frac{1}{z+3}\right) - \left(\frac{1}{z+2} - \frac{1}
{z+3}\right)\\
&= \frac12\frac{1}{z+1} - \frac{1}{z+2} +\frac12\frac{1}{z+3}
\end{align}
$$
Substitute $z$ by $e^{t}$, you obtain the $2^{nd}$ alternate form from WA.
