# Integration by Partial Fractions

• Sep 21st 2008, 04:37 PM
RedBarchetta
Integration by Partial Fractions
$
\int {\frac{{2s + 2}}
{{(s^2 + 1)(s - 1)^3 }}ds}
$

Alright, so I split it up the integrand into partial fractions and try to find the coefficients:

$
\begin{gathered}
\frac{{2s + 2}}
{{(s^2 + 1)(s - 1)^3 }} = \frac{{As + B}}
{{s^2 + 1}} + \frac{C}
{{s - 1}} + \frac{D}
{{(s - 1)^2 }} + \frac{E}
{{(s - 1)^3 }} \hfill \\
2s + 2 = (As + B)(s - 1)^3 + C(s^2 + 1)(s - 1)^2 + D(s^2 + 1)(s - 1) + E(s^2 + 1) \hfill \\
\end{gathered}
$

Plugging in s=1 I get E=2

$
\begin{gathered}
4 = (As + B)(0)^3 + C(s^2 + 1)(0)^2 + D(s^2 + 1)(0) + E(2) \hfill \\
4 = E(2) \Rightarrow E = 2 \hfill \\
\end{gathered}
$

But solving for the other variables gets tricky.....

when s=2

$
\begin{gathered}
6 = (As + B)(1)^3 + C(5)(1)^2 + D(5)(1) + E(5) \hfill \\
2A + B + 5C + 5D = 0 \hfill \\
\end{gathered}
$

when s=-2

$
\begin{gathered}
- 2 = (B - 2a)( - 27) + C(5)( - 3)^2 + D(5)( - 3) + 2(5) \hfill \\
54A + 45C - 15D - 27B + 12 = 0 \hfill \\
\end{gathered}
$

s=-1

$
\begin{gathered}
0 = (B - A)( - 2)^3 + C(2)( - 2)^2 + D(2)( - 2) + 2(2) \hfill \\
0 = 8A - 8B + 8C - 4D + 4 \hfill \\
\end{gathered}
$

....and on and on. How do you even begin to solve these?
• Sep 21st 2008, 04:55 PM
Jhevon
Quote:

Originally Posted by RedBarchetta
$
\int {\frac{{2s + 2}}
{{(s^2 + 1)(s - 1)^3 }}ds}
$

Alright, so I split it up the integrand into partial fractions and try to find the coefficients:

$
\begin{gathered}
\frac{{2s + 2}}
{{(s^2 + 1)(s - 1)^3 }} = \frac{{As + B}}
{{s^2 + 1}} + \frac{C}
{{s - 1}} + \frac{D}
{{(s - 1)^2 }} + \frac{E}
{{(s - 1)^3 }} \hfill \\
2s + 2 = (As + B)(s - 1)^3 + C(s^2 + 1)(s - 1)^2 + D(s^2 + 1)(s - 1) + E(s^2 + 1) \hfill \\
\end{gathered}
$

Plugging in s=1 I get E=2

$
\begin{gathered}
4 = (As + B)(0)^3 + C(s^2 + 1)(0)^2 + D(s^2 + 1)(0) + E(2) \hfill \\
4 = E(2) \Rightarrow E = 2 \hfill \\
\end{gathered}
$

But solving for the other variables gets tricky.....

when s=2

$
\begin{gathered}
6 = (As + B)(1)^3 + C(5)(1)^2 + D(5)(1) + E(5) \hfill \\
2A + B + 5C + 5D = 0 \hfill \\
\end{gathered}
$

when s=-2

$
\begin{gathered}
- 2 = (B - 2a)( - 27) + C(5)( - 3)^2 + D(5)( - 3) + 2(5) \hfill \\
54A + 45C - 15D - 27B + 12 = 0 \hfill \\
\end{gathered}
$

s=-1

$
\begin{gathered}
0 = (B - A)( - 2)^3 + C(2)( - 2)^2 + D(2)( - 2) + 2(2) \hfill \\
0 = 8A - 8B + 8C - 4D + 4 \hfill \\
\end{gathered}
$

....and on and on. How do you even begin to solve these?

it's a bit messy. but i think the more straightforward way is to expand everything and equate coefficients.

but to continue here, you have 4 unknowns left, you need four equations.

plug in s = 0 to get another equation. then you can use the elimination method to solve the system. similar to how you use the elimination method to solve a system of two equations and two unknowns