How does Wolfram get from the first form to the second alternate form?

Question

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)} $

Answer 1

$\dfrac{1}{(e^t + 1)(e^t + 2)(e^t + 3)} = \dfrac{1}{(x + 1)(x + 2)(x + 3)}$, where $x = e^t$.

Let $\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{A}{x + 1} + \dfrac{B}{x + 2} + \dfrac{C}{x + 3}$.

Then $A(x + 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x + 2) = 1$.

For $x = -1$:
$A(-1 + 2)(-1 + 3) = 1 \Rightarrow 2A = 1 \Rightarrow A = 1/2$

For $x = -2$:
$B(-2 + 1)(-2 + 3) = 1 \Rightarrow -B = 1 \Rightarrow B = -1$

For $x = -3$:
$C(-3 + 1)(-3 + 2) = 1 \Rightarrow 2C = 1 \Rightarrow C = 1/2$

$$\boxed{\dfrac{1}{(e^t + 1)(e^t + 2)(e^t + 3)} = \dfrac{1}{2(e^t + 1)} - \dfrac{1}{e^t + 2} + \dfrac{1}{2(e^t + 3)}}$$

Alternative Method
Now, I'll tell you an easier method (which has the same logic as above, and only looks different). Using this method, you can just look at the fraction and write down the partial fractions right away.

Write:
$\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{}{x + 1} + \dfrac{}{x + 2} + \dfrac{}{x + 3}$

1. Ignore (or cover with your finger) the first factor in the denominator:
   $\dfrac{1}{\boxed{\color{white}{(x + 1)}}(x + 2)(x + 3)}$
2. Substitute $x = -1$: $\dfrac{1}{(-1 + 2)(-1 + 3)} = \dfrac{1}{2}$.
3. Write this down as the coefficient of the first fraction:
   $\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{1}{2(x + 1)} + \dfrac{}{x + 2} + \dfrac{}{x + 3}$
4. Similarly for the second factor: $\dfrac{1}{(x + 1)\boxed{\color{white}{(x + 2)}}(x + 3)}$, with $x = -3$ gives $\dfrac{1}{(-2 + 1)(-2 + 3)} = -1$.
5. And for the third factor: $\dfrac{1}{(x + 1)(x + 2)\boxed{\color{white}{(x + 3)}}}$, with $x = -2$ gives $\dfrac{1}{(-3 + 1)(-3 + 2)} = \dfrac{1}{2}$.

In this manner you get $\boxed{\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{1}{2(x + 1)} + \dfrac{-1}{x + 2} + \dfrac{1}{2(x + 3)}}$ in one step.

Answer 2

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.