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)} $

Partial fractions: https://en.wikipedia.org/wiki/Partial_fraction_decomposition#Example_1 — M. Vinay, Jun 15 '14 at 11:23
Already tried partial fractions, so I must've had some calculation erros along the lines, cause the result wasnt' of much help.Thank you! — Victor, Jun 15 '14 at 11:25

M. Vinay · Accepted Answer · 2014-06-15T14:21:35.877

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

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

@Victor You would, because that method is a bit lengthy so you have to be very careful. I've added a simpler method which you can use to get the answer faster, and since it's only step, you can repeat it easily if there's any mistake. — M. Vinay, Jun 15 '14 at 14:23
@Anant Thanks for the link! I didn't know it had a name or how to extend it to the other cases. — M. Vinay, Jun 16 '14 at 02:44
I got to know about the method from this answer, and have been enjoying the exploration ever since :) — Anant, Jun 16 '14 at 15:50

score 5 · Answer 2 · answered Jun 15 '14 at 11:35

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.

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

2 Answers2