# Thread: Partial Fractions -- any tricks or shortcuts?

1. ## 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 $\int\frac{x^2+5}{x^3-x^2+x+3}dx$ to $\int\frac{2}{x^2-2x+3} + \frac{1}{x+1}dx$ without all the $Ax+B$ stuff in between.

2. Hello, cinder!

I was wondering if there are any shortcuts.

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

. . without all the $Ax+B$ stuff in between.
With a prime quadratic factor, there's no way to avoid the $Ax + B$.
. . But there are ways to make life a bit easier.

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

. . I immediately make two fractions for the quadratic.

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

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

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

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

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

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

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

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

3. 
\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

\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}

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