Inverse Laplace Transform Property.
For any constants \(a\) and \(b\) and functions of \(s\) \(F(s)\) and \(G(s)\text{,}\)
- \(\displaystyle \ds \ilap{ a f(t) \pm b g(t) } = a \ilap{f(t)} \pm b \ilap{g(t)}\)
Section D.2 Orphaned Content
This is just a place to put content that was written and may or not be used.
Additional Narrative.
Before diving into specifics, we need to mention the linearity property of the inverse Laplace transform, which will be crucial in this discussion.
| \(t\)-functions | \(s\)-functions | ||
| \(\downarrow\) | \(\downarrow\) | ||
| \(f(t)\) | \(\lap{ f(t) }\) | ||
| \(L_1\) | \(1\) | \(\ds \frac{1}{s}\) | \(s \gt 0\) |
| \(L_2\) | \(e^{at}\) | \(\ds \frac{1}{s-a}\) | \(s \gt a\) |
| \(L_3\) | \(t^n\) | \(\ds \frac{n!}{s^{n+1}}\) | \(s \gt 0\) |
| \(L_4\) | \(\sin (bt)\) | \(\ds \frac{b}{s^2 + b^2}\) | \(s \gt 0\) |
| \(L_5\) | \(\cos(bt)\) | \(\ds \frac{s}{s^2 + b^2}\) | \(s \gt 0\) |
| \(L_6\) | \(t^n e^{at}\) | \(\ds \frac{n!}{(s-a)^{n+1}}\) | \(s \gt a\) |
| \(L_7\) | \(e^{at} \sin(bt)\) | \(\ds \frac{b}{(s-a)^2 + b^2}\) | \(s \gt a\) |
| \(L_8\) | \(e^{at} \cos(bt)\) | \(\ds \frac{s-a}{(s-a)^2 + b^2}\) | \(s \gt a\) |
| Properties | |
| \(P_1\) | \(\ds \lap{ a y(t) \pm b z(t) } = a Y(s) \pm b Z(s) \vphantom{\frac{d}{d}}\) |
| \(P_2\) | \(\ds \lap{ e^{at} y(t) } = Y(s-a) \vphantom{\frac{d}{d}}\) |
| \(P_3\) | \(\ds \lap{ y'(t) } = s\,Y(s) - y(0) \vphantom{\frac{d}{d}}\) |
| \(P_4\) | \(\ds \lap{ y''(t) } = s^2\,Y(s) - s\,y(0) - y'(0) \vphantom{\frac{d}{d}}\) |
| \(P_5\) | \(\ds \lap{ y'''(t) } = s^3\,Y(s) - s^2\,y(0) - s\,y'(0) - y''(0) \vphantom{\frac{d}{d}}\) |
| \(P_6\) | \(\ds \lap{ t^n y(t) } = (-1)^n \frac{d^{(n)}}{ds^{(n)}}Y(s) \vphantom{\frac{d}{d}}\) |
Completing the square is an essential technique for transforming quadratic expressions that don’t directly match a form in the table of common Laplace transforms. However, it’s not the only strategy available. In this section, we’ll explore another important technique: partial fraction decomposition. This method is useful for breaking down complex fractions into simpler components that can each be matched with forms in the Laplace transform table.
Two other forms we may wish to match when we study Laplace transforms are
\begin{equation*}
\frac{b}{(s-a)^2 + b^2} \mbox{ and } \frac{s-a}{(s-a)^2 + b^2}.
\end{equation*}
As before, we work toward making the denominator match first, and then we sort out the numerator second.
Example D.16.
Find \(\ds \lap^{-1}\left\{ \frac{s+9}{s^2 - 2s - 3} \right\}.\)
Solution. Solution
As in the previous examples, the denominator is a second-degree polynomial; therefore it is sensible for us to begin by completing the square in the denominator as we did in the previous two examples.
\begin{align*}
\frac{s+9}{s^2 - 2s - 3} =\amp\ \frac{s+9}{(s^2 - 2s) - 3}\\
=\amp\ \frac{s+9}{(s^2 - 2s + 1) - 3 - 1}\\
=\amp\ \frac{s+9}{(s-1)^2 - 4}\\
=\amp\ \frac{s+9}{(s-1)^2 - 2^2}.
\end{align*}
Take a careful look at the denominator here. It’s really close to matching \(L7\) or \(L8\), but it is not a match because of the negative sign in front of the \(2^2.\) We need to change course when this happens. Another algebraic manipulation that we might consider is a partial fraction decomposition.
We revert to the original expression, but this time, instead of completing the square, we factor the denominator.
\begin{equation*}
\frac{s+9}{s^2 - 2s - 3} = \frac{s+9}{(s-3)(s+1)}.
\end{equation*}
Since each of the factors in the denominator is a distinct linear factor, we know that the form of the partial fraction decomposition is
\begin{equation*}
\frac{s+9}{(s-3)(s+1)} = \frac{A}{s-3} + \frac{B}{s+1}.
\end{equation*}
Our next goal is to determine the coefficients \(A\) and \(B\) in this equations. There are multiple ways to achieve this and we demonstrate just one here. We multiply both sides of the equation by the least common denominator, \((s-3)(s+1)\text{,}\) and then expand and collect like terms, as shown.
\begin{align*}
=\amp\ \frac{A}{s-3}\cdot (s-3)(s+1) + \frac{B}{s+1} \cdot (s-3)(s+1)\\
s+9 =\amp\ A(s+1) + B(s-3)\\
s+9 =\amp\ As + A + Bs - 3B\\
s+9 =\amp\ (As + Bs) + (A - 3B)\\
1s+9 =\amp\ (A+B)s + (A - 3B).
\end{align*}
At this point, we have a polynomial on the left hand side and a polynomial on the right hand side. The only way these can be equal to each other is if the corresponding coefficients are equal. That is, the coefficient on \(s\) on the left hand side is 1, while the coefficient on \(s\) on the right hand side of the equation is \(A+B\text{.}\) Since the polynomials are equal, we know that these are equal. That is, \(A + B = 1.\) Similarly, if we equate the constants, we have \(A - 3B = 9.\) Thus, we have the following system of two linear equations in terms of two unknown variables, \(A\) and \(B\text{.}\)
\begin{align*}
A + B =\amp\ 1\\
A - 3B =\amp\ 9.
\end{align*}
There are many ways to solve such an equation, and you are encouraged to choose the solution technique you like the most. Here we will solve the first equation for \(A\text{,}\) and then substitute into the second equation,
\begin{align*}
A + B =\amp\ 1 \amp A - 3B =\amp\ 9\\
A =\amp\ 1 - B \amp \amp\\
\amp \amp (1 - B) - 3B =\amp\ 9\\
\amp \amp 1 - 4B =\amp\ 9\\
\amp \amp -4B =\amp\ 8\\
\amp \amp B =\amp\ -2\\
A =\amp\ 1 - (-2) \amp \amp\\
=\amp\ 3 \amp \amp
\end{align*}
hence we have
\begin{equation*}
\frac{s+9}{(s-3)(s+1)} = \frac{3}{s-3} + \frac{-2}{s+1}.
\end{equation*}
Remember that our goal is to take the inverse Laplace transform. Our algebraic manipulation was helpful because we took a more complex expression and rewrote it as two simpler fractions. We can now use \(L2\) to find the inverse Laplace transform as follows.
\begin{align*}
=\amp\ \lap^{-1}\left\{ \frac{3}{s-3} + \frac{-2}{s+1} \right\}\\
=\amp\ 3 \lap^{-1}\left\{ \frac{1}{s-3} \right\} - 2 \lap^{-1}\left\{ \frac{1}{s+1} \right\}\\
=\amp\ 3e^{3t} - 2e^{-t}
\end{align*}
Example D.17.
Write \(\ds\frac{s-6}{s^2 - 4s + 29}\) in the form \(\ds\frac{s-a}{(s-a)^2 + b^2}\)
Solution. Solution
As we mentioned earlier, we’ll try to make the denominator match first. Since both of the forms we’re trying to match are of the form \((s-a)^2 + b^2,\) we will complete the square first:
\begin{equation*}
s^2 - 4s + 29 = (s-2)^2 + 25.
\end{equation*}
(If you still don’t remember how to complete the square, look up that primer and do the previous exercises in that section above.)
Let’s rewrite the given expression as follows.
\begin{align*}
\frac{s-6}{s^2 - 4s + 29} =\amp\ \frac{s-6}{(s-2)^2 + 25} \\
=\amp\ \frac{s-6}{(s-2)^2 + 5^2}
\end{align*}
We’ve got the denominator in exactly the right form--it looks just like \((s-a)^2 + b^2,\) with \(a = 2\) and \(b = 5\text{.}\) As in the previous section, once we’ve gotten the denominator in shape, we turn our attention to the numerator. If we look back at the two forms we are trying to match, we see that our expression has an \(s\) in the numerator, so it’s more like \(\ds \frac{s-a}{(s-a)^2 + b^2}\text{.}\) It would be exactly right if we had \(s-a\) in the numerator, which in this case would be \(s-2\text{.}\)
What we do have in the numerator is \(s-6\text{;}\) and we would like it to be \(s-2,\) which means if we added 4, we’d have exactly the right thing. If we want to add 4, we’ll need to compensate by also subtracting 4, like this:
\begin{align*}
\frac{s-6}{(s-2)^2 + 5^2}=\amp\ \frac{s-6+4-4}{(s-2)^2 + 5^2}, \\
=\amp\ \frac{(s-6+4)-4}{(s-2)^2 + 5^2}, \\
=\amp\ \frac{(s-2) - 4}{(s-2)^2 + 5^2}.
\end{align*}
Great! Now we can split this single fraction into two separate fractions:
\begin{equation*}
\frac{(s-2) - 4}{(s-2)^2 + 5^2}
= \underbrace{\frac{s-2}{(s-2)^2 + 5^2}}_{match!} - \frac{4}{(s-2)^2 + 5^2}
\end{equation*}
We’re almost there! The first fraction is a perfect match for the form \(\ds\frac{s-a}{(s-a)^2 + b^2}\) (with \(a = 2\) and \(b = 5\)); but we still have another expression that is not yet a match. The remaining fraction looks like it could eventually match the form \(\ds\frac{b}{(s-a)^2 + b^2}\text{.}\) We would need to have a 5 in the numerator, and we currently have a 4. But we can fix that as we did in the previous section:
\begin{align*}
\frac{4}{(s-2)^2 + 5^2} =\amp\ 4\cdot\frac{1}{(s-2)^2 + 5^2} \\
=\amp\ 4\cdot\frac{5}{5}\cdot
\frac{1}{(s-2)^2 + 5^2} \\
=\amp\ 4\cdot\frac{1}{5}\cdot
\frac{5}{(s-2)^2 + 5^2} \\
=\amp\ \frac{4}{5}\cdot
\frac{5}{(s-2)^2 + 5^2}
\end{align*}
Now let’s put it all together. Here’s everything we did:
\begin{align*}
\frac{s-6}{s^2 - 4s + 9} =\amp\ \frac{s-6}{(s-2)^2 + 25} \mbox{(we completed the square)}\\
=\amp\ \frac{s-6}{(s-2)^2 + 5^2} \mbox{(we rewrote the denominator)}\\
=\amp\ \frac{s-6+4-4}{(s-2)^2 + 5^2} \mbox{(we added and subtracted 4 in the numerator...)}\\
=\amp\ \frac{(s-2) - 4}{(s-2)^2 + 5^2} \mbox{(...so that we could make the numerator match part of the denom)}\\
=\amp\ \frac{s-2}{(s-2)^2 + 5^2} - \frac{4}{(s-2)^2 + 5^2} \mbox{(split into fractions)}\\
=\amp\ \frac{s-2}{(s-2)^2 + 5^2} - 4\cdot\frac{5}{5}\cdot \frac{1}{(s-2)^2 + 5^2} \mbox{(work on the second fraction...)}\\
=\amp\ \frac{s-2}{(s-2)^2 + 5^2} - \frac{4}{5}\cdot \frac{5}{(s-2)^2 + 5^2}\mbox{(...so now it is also in form)}
\end{align*}
As mentioned before, being able to use appropriate algebra to "match" forms is really important when we work with Laplace Transforms. Since it’s really just algebra, now is a great time to practice that skill--so when we are in the middle of studying Laplace Transforms, you can just focus on the "new" stuff.
Now you try some!
- Manipulate each of the following expressions to make it match the form \(\ds \frac{b}{(s-a)^2 + b^2}\) and/or \(\ds \frac{s-a}{(s-a)^2 + b^2}.\)
- \(\ds \frac{s+3}{s^2 - 6s + 10}\)
Solution. Solution
\begin{align*} \frac{s+3}{s^2 - 6s + 10} =\amp\ \frac{s+3}{(s^2 - 6s) + 10}\\ =\amp\ \frac{s+3}{(s^2 - 6s + 9 - 9) + 10}\\ =\amp\ \frac{s+3}{(s^2 - 6s + 9) - 9 + 10}\\ =\amp\ \frac{s+3}{(s - 3)^2 - 9 + 10}\\ =\amp\ \frac{s+3}{(s - 3)^2 + 1}\\ =\amp\ \frac{s+3}{(s - 3)^2 + 1^2}\\ =\amp\ \frac{(s-3+3)+3}{(s - 3)^2 + 1^2}\\ =\amp\ \frac{(s-3)+3+3}{(s - 3)^2 + 1^2}\\ =\amp\ \frac{(s-3)+6}{(s - 3)^2 + 1^2}\\ =\amp\ \frac{s-3}{(s - 3)^2 + 1^2} + \frac{6}{(s - 3)^2 + 1^2}\\ =\amp\ \frac{s-3}{(s - 3)^2 + 1^2} + 6\cdot \frac{1}{(s - 3)^2 + 1^2} \\ =\amp\ \frac{s - a}{(s - a)^2 +b^2} + 6\cdot \frac{b}{(s - a)^2 +b^2}, \mbox{ where and } \end{align*}Answer. Answer
\begin{equation*} \frac{s-3}{(s - 3)^2 + 1^2} + 6\cdot \frac{1}{(s - 3)^2 + 1^2} \end{equation*} - \(\ds \frac{s}{s^2 - 8s + 25}\)
Solution. Solution
\begin{align*} \frac{s}{s^2 - 8s + 25} =\amp\ \frac{s}{(s^2 - 8s) + 25}\\ =\amp\ \frac{s}{(s^2 - 8s + 16 - 16) + 25}\\ =\amp\ \frac{s}{(s^2 - 8s + 16) - 16 + 25}\\ =\amp\ \frac{s}{(s - 4)^2 - 16 + 25}\\ =\amp\ \frac{s}{(s - 4)^2 +9}\\ =\amp\ \frac{s}{(s - 4)^2 +3^2}\\ =\amp\ \frac{(s - 4) + 4}{(s - 4)^2 +3^2}\\ =\amp\ \frac{s - 4}{(s - 4)^2 +3^2} + \frac{4}{(s - 4)^2 +3^2}\\ =\amp\ \frac{s - 4}{(s - 4)^2 +3^2} + 4\cdot \frac{1}{(s - 4)^2 +3^2}\\ =\amp\ \frac{s - 4}{(s - 4)^2 +3^2} + \frac{3}{3}\cdot 4\cdot \frac{1}{(s - 4)^2 +3^2}\\ =\amp\ \frac{s - 4}{(s - 4)^2 +3^2} + \frac{1}{3}\cdot 4\cdot \frac{3}{(s - 4)^2 +3^2}\\ =\amp\ \frac{s - 4}{(s - 4)^2 +3^2} + \frac{4}{3}\cdot \frac{3}{(s - 4)^2 +3^2}\\ =\amp\ \frac{s - a}{(s - a)^2 +b^2} + \frac{4}{3}\cdot \frac{b}{(s - a)^2 +b^2}, \end{align*}Answer. Answer
\begin{equation*} \frac{s - 4}{(s - 4)^2 +3^2} + \frac{4}{3}\cdot \frac{3}{(s - 4)^2 +3^2} \end{equation*} - \(\ds \frac{s}{s^2 + 8s + 25}\)
Solution. Solution
\begin{align*} \frac{s}{s^2 + 8s + 25} =\amp\ \frac{s}{(s^2 + 8s) + 25}\\ =\amp\ \frac{s}{(s^2 + 8s + 16 - 16) + 25}\\ =\amp\ \frac{s}{(s^2 + 8s + 16) - 16 + 25}\\ =\amp\ \frac{s}{(s + 4)^2 - 16 + 25}\\ =\amp\ \frac{s}{(s + 4)^2 +9}\\ =\amp\ \frac{s}{(s + 4)^2 +3^2}\\ =\amp\ \frac{(s + 4) - 4}{(s + 4)^2 +3^2}\\ =\amp\ \frac{s + 4}{(s + 4)^2 +3^2} - \frac{4}{(s + 4)^2 +3^2}\\ =\amp\ \frac{s + 4}{(s + 4)^2 +3^2} - 4\cdot \frac{1}{(s + 4)^2 +3^2}\\ =\amp\ \frac{s + 4}{(s + 4)^2 +3^2} - \frac{3}{3}\cdot 4\cdot \frac{1}{(s + 4)^2 +3^2}\\ =\amp\ \frac{s + 4}{(s + 4)^2 +3^2} - \frac{1}{3}\cdot 4\cdot \frac{3}{(s + 4)^2 +3^2}\\ =\amp\ \frac{s + 4}{(s + 4)^2 +3^2} - \frac{4}{3}\cdot \frac{3}{(s + 4)^2 +3^2}\\ =\amp\ \frac{s - a}{(s - a)^2 +b^2} - \frac{4}{3}\cdot \frac{b}{(s - a)^2 +b^2}, \end{align*}Answer. Answer
\begin{equation*} \frac{s + 4}{(s + 4)^2 +3^2} - \frac{4}{3}\cdot \frac{3}{(s + 4)^2 +3^2} \end{equation*} - \(\ds \frac{s-5}{s^2 +12s + 72}\)
Solution. Solution
\begin{align*} \frac{s-5}{s^2 +12s + 72} =\amp\ \frac{s-5}{(s^2 +12s) + 72}\\ =\amp\ \frac{s-5}{(s^2 +12s + 36 - 36) + 72}\\ =\amp\ \frac{s-5}{(s^2 +12s + 36) - 36 + 72}\\ =\amp\ \frac{s-5}{(s + 6)^2 - 36 + 72}\\ =\amp\ \frac{s-5}{(s + 6)^2 + 36}\\ =\amp\ \frac{s-5}{(s + 6)^2 + 6^2}\\ =\amp\ \frac{(s+6-6)-5}{(s + 6)^2 + 6^2}\\ =\amp\ \frac{(s+6)-6-5}{(s + 6)^2 + 6^2}\\ =\amp\ \frac{(s+6)-11}{(s + 6)^2 + 6^2}\\ =\amp\ \frac{s+6}{(s + 6)^2 + 6^2} + \frac{-11}{(s + 6)^2 + 6^2}\\ =\amp\ \frac{s+6}{(s + 6)^2 + 6^2} -11\cdot \frac{1}{(s + 6)^2 + 6^2}\\ =\amp\ \frac{s+6}{(s + 6)^2 + 6^2} - \frac{6}{6} \cdot 11\cdot \frac{1}{(s + 6)^2 + 6^2}\\ =\amp\ \frac{s+6}{(s + 6)^2 + 6^2} - \frac{1}{6} \cdot 11\cdot \frac{6}{(s + 6)^2 + 6^2}\\ =\amp\ \frac{s+6}{(s + 6)^2 + 6^2} - \frac{11}{6}\cdot \frac{6}{(s + 6)^2 + 6^2}\\ =\amp\ \frac{s - a}{(s - a)^2 +b^2} - \frac{11}{6}\cdot \frac{b}{(s - a)^2 +b^2}, \end{align*}Answer. Answer
\begin{equation*} \frac{s+6}{(s + 6)^2 + 6^2} - \frac{11}{6}\cdot \frac{6}{(s + 6)^2 + 6^2} \end{equation*}
- Now that’s you’ve been practicing manipulating expressions to get them to match a particular, specific form, it’s time to practice picking out which form an expression is most like. Find the form(s) on the right that best match the expression on the left. It may not be a perfect match, but could be manipulated (as we did above) to get into the correct form(s).
- (EXAMPLE) \(\ds \frac{1}{(s-3)^5}\)
Answer.
- \(\ds \frac{12}{s^2 + 15}\)%D
- \(\ds \frac{7}{1-s}\) %B
- \(\ds \frac{-1}{s}\) %A
- \(\ds \frac{s-1}{s^2 + 4s + 1}\) %forms G and H
- \(\ds \frac{7}{2s^2}\) %C
- \(\ds \frac{s+2}{s^2 + 9}\) %forms D and E
- \(\displaystyle \ds \frac{1}{s}\)
- \(\displaystyle \ds \frac{1}{s-a}\)
- \(\displaystyle \ds \frac{n!}{s^{n+1}}\)
- \(\displaystyle \ds \frac{b}{s^2 + b^2}\)
- \(\displaystyle \ds \frac{s}{s^2 + b^2}\)
- \(\displaystyle \ds \frac{n!}{(s-a)^{n+1}}\)
- \(\displaystyle \ds \frac{b}{(s-a)^2 + b^2}\)
- \(\displaystyle \ds \frac{s-a}{(s-a)^2 + b^2}\)
Solution. Solution
- This best matches form F.
- This best matches form D.
- This best matches form B.
- This best matches form A.
- This best matches form forms G and H.
- This best matches form C.
- This best matches form forms D and E.
If you have a rational function where the denominator is of higher degree, then partial fraction decomposition should be used to break the single fraction into several simpler fractions.
Example D.18.
Find the inverse Laplace transform of
\begin{equation*}
R(s) = \frac{4s^3 - 13s^2 + 74s + 27}{s^4 - 4s^3 + 14s^2 + 44s + 25} = \frac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2}.
\end{equation*}
Solution. Solution
Note that as the quadratic term in the denominator does not factor, the denominator contains an irreducible quadratic factor and a repeated linear factor. We’ll proceed by simplifying this complicated fraction with a Partial Fraction Decomposition of the form
\begin{equation*}
\frac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2}
= \frac{As + B}{s^2 - 6s + 25} + \frac{C}{s+1} + \frac{D}{(s+1)^2}.
\end{equation*}
You may also consider using technology to find a partial fraction decomposition. You should get
\begin{equation*}
\frac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2} = \frac{s + 2}{s^2 - 6s + 25} + \frac{3}{s+1} + \frac{-2}{(s+1)^2}.
\end{equation*}
With the partial fraction decomposition in hand, we are prepared to take the inverse Laplace transform, using the same types of algebraic manipulations demonstrated in the previous examples.
\begin{align*}
r(t) =\amp\ \lap^{-1}\left\{ R(s) \right\}\\
=\amp\ \lap^{-1}\left\{ \frac{s + 2}{s^2 - 6s + 25} + \frac{3}{s+1}
+ \frac{-2}{(s+1)^2} \right\} \\
=\amp\ \lap^{-1}\left\{ \frac{s + 2}{s^2 - 6s + 25}\right\}
+ 3\cdot \lap^{-1}\left\{ \frac{1}{s+1} \right\}
- 2\cdot \lap^{-1}\left\{\frac{1}{(s+1)^2} \right\} \\
=\amp\ \lap^{-1}\left\{ \frac{s + 2}{(s-3)^2 + 4^2}\right\}
+ 3e^{-t} - 2te^{-t}\\
=\amp\ \lap^{-1}\left\{ \frac{(s - 3) + 5}{(s-3)^2 + 4^2}\right\}
+ 3e^{-t} - 2te^{-t}\\
=\amp\ \lap^{-1}\left\{ \frac{s - 3}{(s-3)^2 + 4^2}\right\}
+ 5 \lap^{-1}\left\{ \frac{1}{(s-3)^2 + 4^2}\right\}
+ 3e^{-t} - 2te^{-t}\\
=\amp\ e^{3t}\cos(4t)
+ \frac{5}{4}\cdot \lap^{-1}\left\{ \frac{4}{(s-3)^2 + 4^2}\right\}
+ 3e^{-t} - 2te^{-t}\\
=\amp\ e^{3t}\cos(4t)
+ \frac{5}{4}e^{3t}\sin(4t)
+ 3e^{-t} - 2te^{-t}.
\end{align*}
In summary, when we want to take the inverse Laplace transform of a rational function with a second-degree polynomial in the denominator, we may complete the square or we may do a partial fraction decomposition. How will we know which is appropriate? Here are a few guidelines for you to consider.
- Does the denominator factor in an obvious way? If so, factor the denominator and do a partial fraction decomposition if necessary.
- If the denominator does not factor in an obvious way, try completing the square.
- If instead you end up with subtraction outside the parentheses, as in \((s - a)^2 - b^2,\) then you should factor and do a partial fraction decomposition. You may consider using the quadratic formula if the factorization is not obvious to you.
- If you end up with no terms outside the parentheses, as in \((s - a)^2,\) then use \(L6\).
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