Is there a systematic method at use to turn something like $\frac{1}{2+u}\cdot\frac{1}{1+u^2}$ into $\frac{1}{2+u}-\frac{u-2}{1+u^2}$?

Question

Is there some kind of method/trick/strategy/etc used to turn

$$\frac{1}{2+u}\cdot\frac{1}{1+u^2}$$

into

$$\frac{1}{5}\left [\frac{1}{2+u}-\frac{u-2}{1+u^2}\right ]$$

?

I run into such problems while integrating.

For the example above, the integral was

$$\int\frac{dx}{2+\tan{x}}$$

In Spivak's Calculus, such algebra is done out of the blue. Is it just either cleverness or trial and error or a mix of both is there some method?

For the record, Spivak does say "there is no substitute for cleverness" at the beginning of this multi-item problem of integration. I just want to confirm that in fact it is as difficult as it seems to pull such decompositions out of thin air.

Answer 1

The partial fraction decomposition is introduced on page 374 of Spivak's "Calculus" which you refer to in the question. Here is this example.

Let $$ \begin{align} \frac{1}{2+u}\cdot\frac{1}{1+u^2} = {a\over2+u}+{bu+c\over 1+u^2} \end{align} $$ then multiplying out gives $$ {(2+u)(bu+c)+a(1+u^2)\over (2+u)(1+u^2) } ={(a+b)u^2+(2b+c)u+2c+a \over (2+u)(1+u^2)}. $$ Since there is no $u^2$ on the top we must have $a=-b$, similarly for $u$ gives $c=-2b=2a$, then $2c+a=1$ gives $5a=1$, so $a=1/5$, $b=-1/5$, $c=2/5$.

Answer 2

You may avoid a lengthy partial fraction procedure by decomposing the integrand of the general form below as follows

\begin{align}\frac{1}{(a+u)(1+u^2)} =& \ \frac1{a^2+1}\cdot \frac{(1+u^2)+(a^2-u^2)}{(a+u)(1+u^2)}\\ =&\ \frac1{a^2+1}\cdot \bigg( \frac{1}{a+u}+ \frac{a-u}{1+u^2}\bigg) \end{align}

Answer 3

We can use complex numbers $$f(u):={1\over (u+2)(u+i)(u-i)}={a\over u+2}+{b\over u+i}+{\bar{b}\over u-i}$$ Then $$a=\lim_{u\to 2}f(u)(u+2)={1\over (2+i)(2-i)}={1\over 5}$$ $$b=\lim_{u\to -i}f(u)(u+i)={1\over (2-i)(-i-i)}={ 2i-1\over 10}$$ Observe that $${b\over u+i}+{\bar{b}\over u-i}={2\Re b\,u -2i{\Im b}\over u^2+1}$$ hence $$ f(u)= {1\over 5}{1\over u+2}-{1\over 5} {u-2\over u^2+1}$$