# Partial Fractions -- any tricks or shortcuts?

• Jul 10th 2007, 05:41 PM
cinder
Partial Fractions -- any tricks or shortcuts?
Today in class we did partial fractions. I understand the process, but I was wondering if there are any shortcuts.

For example: perhaps a shortcut way to go from $\displaystyle \int\frac{x^2+5}{x^3-x^2+x+3}dx$ to $\displaystyle \int\frac{2}{x^2-2x+3} + \frac{1}{x+1}dx$ without all the $\displaystyle Ax+B$ stuff in between.
• Jul 10th 2007, 06:28 PM
Soroban
Hello, cinder!

Quote:

I was wondering if there are any shortcuts.

Perhaps a way to go from $\displaystyle \int\frac{x^2+5}{x^3-x^2+x+3}\,dx$ .to .$\displaystyle \int\left(\frac{2}{x^2-2x+3} + \frac{1}{x+1}\right)dx$

. . without all the $\displaystyle Ax+B$ stuff in between.

With a prime quadratic factor, there's no way to avoid the $\displaystyle Ax + B$.
. . But there are ways to make life a bit easier.

We have: .$\displaystyle \frac{x^2 + 5}{(x+1)(x^2-2x+3)} \;=\;\frac{A}{x+1} + \frac{Bx}{x^2-2x+3} + \frac{C}{x^2-2x+3}$

. . I immediately make two fractions for the quadratic.

Clear denominators: .$\displaystyle x^2 + 5 \;=\;A(x^2-2x+3) + Bx(x+1) + C(x + 1)$

Let $\displaystyle x = \text{-}1\!:\;\;6\;=\;A(6) + B(0) + C(0)\quad\Rightarrow\quad\boxed{A = 1}$

. . I immediately replace $\displaystyle A$ with $\displaystyle 1\!:\;\;x^2 + 5 \;=\;(x^2 - 2x + 3) + Bx(x + 1) + C(x + 1)$

Let $\displaystyle x = 0\!:\;\;5 \;=\;(3) + B(0) + C(1)\quad\Rightarrow\quad\boxed{ C = 2}$

. . Replace $\displaystyle C$ with $\displaystyle 2\!:\;\;x^2 + 5\;=\;(x^2-2x+3) + Bx(x+1) + 2(x+1)$

Let $\displaystyle x = 1\!:\;\;6 \;=\;(2) + B(2) + (4)\quad\Rightarrow\quad\boxed{B = 0}$

Therefore: .$\displaystyle \frac{x^2+5}{(x+1)(x^2-2x+3)} \;=\;\frac{1}{x+1} + \frac{2}{x^2-2x+3}$

. . . . . with a minimum of pain and aggravation.

• Jul 10th 2007, 06:41 PM
Krizalid
\displaystyle \begin{aligned} x^3 - x^2 + x + 3 &= x^3 - (2x^2 - x^2 ) + (3x - 2x) + 3\\ &= (x^3 + x^2 ) - (2x^2 + 2x) + (3x + 3)\\ &= x^2 (x + 1) - 2x(x + 1) + 3(x + 1)\\ &= (x + 1)(x^2 - 2x + 3) \end{aligned}

Therefore

\displaystyle \begin{aligned} \int {\frac{{x^2 + 5}} {{x^3 - x^2 + x + 3}}~dx} &= \int {\frac{{x^2 + 5}} {{(x + 1)(x^2 - 2x + 3)}}~dx}\\ &= \int {\frac{{2(x + 1) + (x^2 - 2x + 3)}} {{(x + 1)(x^2 - 2x + 3)}}~dx}\\ &= \int {\left( {\frac{2} {{x^2 - 2x + 3}} + \frac{1} {{x + 1}}} \right)~dx} \end{aligned}